sample problem solving in kinetic energy

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Potential And Kinetic Energy Example Problem – Work and Energy Examples

Potential energy is energy attributed to an object by virtue of its position. When the position is changed, the total energy remains unchanged but some potential energy gets converted into kinetic energy . The frictionless roller coaster is a classic potential and kinetic energy example problem.

The roller coaster problem shows how to use the conservation of energy to find the velocity or position or a cart on a frictionless track with different heights. The total energy of the cart is expressed as a sum of its gravitational potential energy and kinetic energy. This total energy remains constant across the length of the track.

Potential And Kinetic Energy Example Problem

Rollercoaster Diagram for Conservation of Energy Example Problem

A cart travels along a frictionless roller coaster track. At point A, the cart is 10 m above the ground and traveling at 2 m/s. A) What is the velocity at point B when the cart reaches the ground? B) What is the velocity of the cart at point C when the cart reaches a height of 3 m? C) What is the maximum height the cart can reach before the cart stops?

The total energy of the cart is expressed by the sum of its potential energy and its kinetic energy.

Potential energy of an object in a gravitational field is expressed by the formula

where PE is the potential energy m is the mass of the object g is the acceleration due to gravity = 9.8 m/s 2 h is the height above the measured surface.

Kinetic energy is the energy of the object in motion. It is expressed by the formula

KE = ½mv 2

where KE is the kinetic energy m is the mass of the object v is the velocity of the object.

The total energy of the system is conserved at any point of the system. The total energy is the sum of the potential energy and the kinetic energy.

Total E = KE + PE

To find the velocity or position, we need to find this total energy. At point A, we know both the velocity and the position of the cart.

Total E = KE + PE Total E = ½mv 2 + mgh Total E = ½m(2 m/s) 2 + m(9.8 m/s 2 )(10 m) Total E = ½m(4 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(2 m 2 /s 2 ) + m(98 m 2 /s 2 ) Total E = m(100 m 2 /s 2 )

We can leave the mass value as it appears for now. As we complete each part, you will see what happens to this variable.

The cart is at ground level at point B, so h = 0 m.

Total E = ½mv 2 + mgh Total E = ½mv 2 + mg(0 m) Total E = ½mv 2

All of the energy at this point is kinetic energy. Since total energy is conserved, the total energy at point B is the same as the total energy at point A.

Total E at A = Total Energy at B m(100 m 2 /s 2 ) = ½mv 2

Divide both sides by m 100 m 2 /s 2 = ½v 2

Multiply both sides by 2 200 m 2 /s 2 = v 2

v = 14.1 m/s

The velocity at point B is 14.1 m/s.

At point C, we know only a value for h (h = 3 m).

Total E = ½mv 2 + mgh Total E = ½mv 2 + mg(3 m)

As before, the total energy is conserved. Total energy at A = total energy at C.

m(100 m 2 /s 2 ) = ½mv 2 + m(9.8 m/s 2 )(3 m) m(100 m 2 /s 2 ) = ½mv 2 + m(29.4 m 2 /s 2 )

Divide both sides by m

100 m 2 /s 2 = ½v 2 + 29.4 m 2 /s 2 ½v 2 = (100 – 29.4) m 2 /s 2 ½v 2 = 70.6 m 2 /s 2 v 2 = 141.2 m 2 /s 2 v = 11.9 m/s

The velocity at point C is 11.9 m/s.

The cart will reach its maximum height when the cart stops or v = 0 m/s.

Total E = ½mv 2 + mgh Total E = ½m(0 m/s) 2 + mgh Total E = mgh

Since total energy is conserved, the total energy at point A is the same as the total energy at point D.

m(100 m 2 /s 2 ) = mgh

100 m 2 /s 2 = gh

100 m 2 /s 2 = (9.8 m/s 2 ) h

The maximum height of the cart is 10.2 m.

A) The velocity of the cart at ground level is 14.1 m/s. B) The velocity of the cart at a height of 3 m is 11.9 m/s. C) The maximum height of the cart is 10.2 m.

This type of problem has one main key point: total energy is conserved at all points of the system. If you know the total energy at one point, you know the total energy at all points.

Kinetic Energy Formula | Problems (With Solutions)

Kinetic energy (KE) is half times the product of mass (m), and the square of velocity (v). Using the formula of kinetic energy: KE = ½ × m v 2 , the value of kinetic energy of an object can be calculated.

Let’s solve some problems based on this formula, so you’ll get a clear idea.

Kinetic Energy Practice Problems

Problem 1: A ball of mass 3 kg is moving with the velocity of 6 m/s. Calculate the kinetic energy of a ball.

Solution: Given data: Mass of a ball, m = 3 kg Velocity of a ball, v = 6 m/s Kinetic energy of a ball, KE = ? Using the formula of kinetic energy, KE = ½ × m v 2 KE = ½ × 3 × (6) 2 KE = ½ × 3 × 36 KE = 54 J Therefore, the kinetic energy of a ball is 54 J .

Problem 2: What is the kinetic energy of a 6 kg bicycle running with the velocity of 12 m/s?

Solution: Given data: Mass of a bicycle, m = 6 kg Velocity of a bicycle, v = 12 m/s Kinetic energy of a bicycle, KE = ? Using the formula of kinetic energy, KE = ½ × m v 2 KE = ½ × 6 × (12) 2 KE = 3 × 144 KE = 432 J Therefore, the kinetic energy of a bicycle is 432 J .

Problem 3: Calculate the kinetic energy of a 800 kg car moving with the velocity of 3 m/s.

Solution: Given data: Kinetic energy of a car, KE = ? Mass of a car, m = 800 kg Velocity of a car, v = 3 m/s Using the formula of kinetic energy, KE = ½ × m v 2 KE = ½ × 800 × (3) 2 KE = 400 × 9 KE = 3600 J Therefore, the kinetic energy of a car is 3600 J .

Problem 4: A 400 kg rail wagon has a velocity of 7 m/s. What is the kinetic energy of a rail wagon?

Solution: Given data: Mass of a rail wagon, m = 400 kg Velocity of a rail wagon, v = 7 m/s Kinetic energy of a rail wagon, KE = ? Using the formula of kinetic energy, KE = ½ × m v 2 KE = ½ × 400 × (7) 2 KE = 200 × 49 KE = 9800 J Therefore, the kinetic energy of a rail wagon is 9800 J .

How to Calculate Kinetic Energy

Last Updated: September 22, 2022 References

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There are two basic forms of energy: potential and kinetic energy. Potential energy is the energy an object has relative to the position of another object. [1] X Research source For example, if you are at the top of a hill, you have more potential energy than if you are at the bottom of the hill. Kinetic energy is the energy an object has when it is in motion. [2] X Research source Kinetic energy can be due to vibration, rotation, or translation (movement from one place to another). [3] X Research source The kinetic energy of an object can easily be determined by an equation using the mass and velocity of that object. [4] X Research source

Understanding Kinetic Energy

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Your answer should always be stated in joules (J), which is the standard unit of measurement for kinetic energy. It is equivalent to 1 kg * m 2 /s 2 .

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Calculating Kinetic Energy

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Using Kinetic Energy to Find Velocity or Mass

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About This Article

To calculate kinetic energy, write out a formula where kinetic energy is equal to 0.5 times mass times velocity squared. Add in the value for the mass of the object, then the velocity with which it is moving. Solve for the unknown variable. Your answer should be stated in joules, or J. If you want to learn how to solve velocity or mass using kinetic energy, keep reading the article! Did this summary help you? Yes No

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Kinetic energy problems

When solving kinetic energy problems, you may be asked to find 3 variables. These variables are the kinetic energy, the mass, or the speed.

Problem # 1:

Suppose a car has 3000 Joules of kinetic energy. What will be its kinetic energy if the speed is doubled? What if the speed is tripled?

We already proved in kinetic energy lesson that whenever the speed is doubled, the kinetic energy is quadrupled or four times as big.

4 × 3000 = 12000

Therefore, the kinetic energy is going to be 12000 joules.

Let v be the speed of a moving object. Let speed = 3v after the speed is tripled.

9 × 3000 = 27000

Therefore, the kinetic energy is going to be 27000 joules.

Problem # 2:

Calculate the kinetic energy of a 10 kg object moving with a speed of 5 m/s. Calculate the kinetic energy again when the speed is doubled.

Tricky kinetic energy problems

Problem # 3:

Suppose a rat and a rhino are running with the same kinetic energy. Which one do you think is going faster?

The only tricky and hard part is to use the kinetic energy formula to solve for v.

Multiply both sides by 2

Problem # 4:

The kinetic energy of an object is 8 times bigger than the mass. Is it possible to get speed of the object?

Think carefully and try to solve this problem yourself.

Potential energy

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Kinetic energy problem solving examples

Here, we will be discussing about Kinetic energy problem solving examples.

Kinetic Energy Examples

Energy Problems. The Kinetic Energy (Ek) of an object depends on both its mass (m) (You can be asked to solve problems similar to the lift example.

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Examples of Kinetic Energy Problems.

Kinetic Energy Example: What is the KE of a 1500 kg car going at suburban speed of 14 m/s (about 50 km/h or 30 mph)? Example: The same car is now going at

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Kinetic Energy Formula

Kinetic energy problem solving examples

Best of all, Kinetic energy problem solving examples is free to use, so there's no sense not to give it a try!

Kinetic Energy Examples

Problem 1: A car is travelling at a velocity of 10 m/s and it has a mass of 250 Kg. Compute its Kinetic energy? Answer: Given: Mass of the body

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Kinetic Energy Formula

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Practice Problems on Kinetic Energy

Kinetic Energy Practice Problems. 1. What is the Kinetic Energy of a 150 kg object that is moving with a speed of 15 m/s? KE = mv2.

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Kinetic Energy Examples

A series of free GCSE/IGCSE Physics Notes and Lessons .

In these lessons, we will

Kinetic Energy

The following diagram shows the formula for kinetic energy. Scroll down the page for more examples and solutions on how to use the formula.

Kinetic energy is the energy stored in moving objects. Stationary objects have no kinetic energy.

E k = 0.5 × m × v 2

Kinetic Energy - IGCSE Physics

Calculations using the kinetic energy formula.

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Rotational Kinetic Energy - Problem Solving

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A variety of problems can be framed on the concept of rotational kinetic energy. The problems can involve the following concepts,

1) Kinetic energy of rigid body under pure translation or pure rotation or in general plane motion. 2) Work done by torque and its relation with rotational kinetic energy in case of fixed axis rotation. 3) Conservation of mechanical energy.

Rotational kinetic energy

Work-kinetic theory for rotation, conservation of energy.

A rod of mass 'M' and length 'L' is rotating about an axis passing through its end and perpendicular to its length. If the angular velocity of rotation at an instant is \(\omega \) then find its kinetic energy. As the axis of rotation of the rod is fixed thus the rod is in pure rotation and its rotational kinetic energy is given by \[KE = \frac{1}{2}{I_{rot}}{\omega ^2}\] here, \({I_{rot}}\) is the moment of inertia of rod about the axis of rotation, which is \[{I_{rod\,about\,end}} = \frac{{M{L^2}}}{3}\] Thus the kinetic energy is given by \[KE = \frac{1}{2}\frac{{M{L^2}}}{3}{\omega ^2} = \frac{{M{L^2}{\omega ^2}}}{6}\]

A uniform hoop (ring) of mass M and radius R is rolling without slipping on a horizontal ground with its center having velocity 'v'. Find the kinetic energy of the hoop. The ring is in general plane motion, thus its motion can be thought as the combination of pure translation of the center of mass and pure rotation about the center of mass. The kinetic energy of the hoop will be written as, \[KE = \frac{1}{2}MV_{_{cm}}^2 + \frac{1}{2}{I_{cm}}{\omega ^2}\] Here \({V_{cm}}\) is the speed of the center of mass and \({I_{cm}}\) is the moment of inertia about an axis passing through its center of mass and perpendicular to the plane of the hoop. \[{I_{cm,hoop}} = M{R^2}\] \[KE = \frac{1}{2}Mv_{}^2 + \frac{1}{2}M{R^2}{\omega ^2}\] For pure rolling motion (rolling without slipping) \[v = r\omega \] \[KE = \frac{1}{2}Mv_{}^2 + \frac{1}{2}M{v^2} = M{v^2}\]

A wheel of mass 'm' and radius 'R' is rolling on a level road at a linear speed 'V'. The kinetic energy of the upper right quarter part of the wheel will be:

Details and assumptions:

Consider the wheel to be of the form of a disc.

A pulley of mass M has a thread wound around it tightly, as shown in the diagram. A constant force starts acting on the open end of the thread. If the pulley is initially at rest, then find the angular speed of the pulley as a function of angle rotated by the pulley. There are three forces acting on the pulley 1) Force by thread 2) Gravitational force acting on the center of mass of the pulley 3) Force by hinge Torque of hinge force and gravitational force about the center of the pulley is zero as they pass through the center itself. Thus, the net torque about the center of the pulley equals \[\vec \tau = \vec r \times \vec F\] \[\tau = FR\] The torque is constant, thus the net work done by the torque on rotating the pulley by an angle \(\theta \) equals, \[W = FR\theta \] The work done by the torque goes into increasing the rotational kinetic energy of the pulley, Thus, according to the work energy theorem for rotation, \[\begin{array}{l} {W_\tau } = \Delta KE\\ FR\theta = \frac{1}{2}I({\omega ^2} - {0^2}) \end{array}\] A pulley can be considered as a disc, thus the moment of inertia \(I = \frac{{M{R^2}}}{2}\) \[\omega = \sqrt {\frac{{4F\theta }}{{MR}}} \]

A sphere is released from the top of a rough inclined plane. The friction is sufficient so that the sphere rolls without slipping. Mass of the sphere is M and radius is R. The height of the center of the sphere from ground is h. Find the speed of the center of the sphere as it reaches the bottom of the sphere. In case of pure rolling on the fixed inclined plane, the point of contact remains at rest and work done by friction is zero. If sphere and earth are taken into one system, then the gravitational force becomes internal force. Other external force, Normal reaction is perpendicular to the direction of motion, thus will not do any work. Thus, no external force or non conservative forces are doing work, and mechanical energy of the system can be conserved. When the ball reaches the bottom of the inclined plane, then its center is moving with speed 'v' and the ball is also rotating about its center of mass with angular velocity \(\omega \). In pure rolling motion, v and \(\omega \) are related as \[v = R\omega \] As the ball comes down the potential energy decreases and therefore kinetic energy increases. The center of ball decends by 'h-R', Loss in potential energy = gain in kinetic energy \[Mg(h - R) = \frac{1}{2}M{v^2} + \frac{1}{2}{I_{cm}}{\omega ^2}\] Moment of inertia of sphere about an axis passing through the center of mass equals \[{I_{cm,sphere}} = \frac{2}{5}M{R^2}\] Therefore, \[\begin{array}{l} g(h - R) = \frac{7}{{10}}{v^2}\\ v = \sqrt {\frac{{10g(h - R)}}{7}} \end{array}\]

A rod of mass \(M\) and length \(L\) is hinged at its end and is in horizontal position initially. It is then released to fall under gravity. Find the angular speed of rotation of rod when the rod becomes vertical.

A meter stick is pivoted about its horizontal axis through its center, has a body of mass 2 kg attached to one end and a body of mass 1 kg attached to the other. The mass of the meter stick can be neglected. The system is released from rest with the stick horizontal. what is the velocity of each body in m/s as the stick swings through a vertical position? Take g = 9.8 m/s^2

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Kinetic energy problem solving examples

Sample Problems. Question 1: A ball has a mass of 2Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it.

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Potential and Kinetic Energy

Kinetic Energy KE = 0.5*m*v KE = (0.5) * (625 kg) * (18.3 m/s) KE = 1.05 x105 Joules. 2. If the speed is doubled, then the KE is quadrupled. Thus, KE = 4 *

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Kinetic Energy Practice Problems

When solving kinetic energy problems, you may be asked to find 3 variables. These variables are the kinetic energy, the mass, or the speed. Problem # 1:.

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Examples of Kinetic Energy Problems.

Learn how to solve kinetic energy problems, and see examples that walk through sample problems step-by-step for you to improve your math knowledge and

Practice Problems on Kinetic Energy

Physics Formulas

Kinetic Energy Formula

Kinetic energy is the energy possessed by a body due to its motion. Kinetic Energy Formula is articulated as

mass of the body = m ,

the velocity with which the body is travelling is v .

The Kinetic energy is articulated in Kgm 2 /s 2

Kinetic energy formula is used to compute the mass, velocity or kinetic energy of the body if any of the two numerics are given.

Kinetic Energy Solved Examples

Underneath are questions on Kinetic energy which aids one to understand where they can use these questions.

Problem 1: A car is travelling at a velocity of 10 m/s and it has a mass of 250 Kg. Compute its Kinetic energy? Answer:

Given: Mass of the body m = 250 Kg, Velocity v = 10 m/s,

Kinetic energy is given by

=12500 kgm 2 s 2

Problem 2: A man is transporting a trolley of mass 6 Kg and having Kinetic energy of 40 J. Compute its Velocity with which he is running? Answer:

Given: Mass, m = 6 Kg Kinetic energy K.E = 60 J

The man is running with the velocity of 3.65m/s

the maximum speed of “tsunami” waves is 100m/s .calculate the maximum kinetic energy of every kilogram of water in these “tsunami “.

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Potential and Kinetic Energy

Energy is the capacity to do work .

The unit of energy is J (Joule) which is also kg m 2 /s 2 (kilogram meter squared per second squared)

Energy can be in many forms! Here we look at Potential Energy (PE) and Kinetic Energy (KE).

Potential Energy and Kinetic Energy

Potential energy (PE) is stored energy due to position or state

Kinetic energy (KE) is energy of motion

From PE to KE

For a good example of PE and KE have a play with a pendulum .

Gravitational Potential Energy

When the PE is due to an objects height then:

PE due to gravity = m g h

Example: This 2 kg hammer is 0.4 m up. What is it's PE?

Kinetic energy.

The formula is:

KE = ½ m v 2

Example: What is the KE of a 1500 kg car going at suburban speed of 14 m/s (about 50 km/h or 30 mph)?

KE = ½ × 1500 kg × (14 m/s) 2

KE = 147,000 kg m 2 /s 2

KE = 147 kJ

Let's double the speed!

Example: The same car is now going at highway speed of 28 m/s (about 100 km/h or 60 mph)?

KE = ½ × 1500 kg × (28 m/s) 2

KE = 588,000 kg m 2 /s 2

KE = 588 kJ

Wow! that is a big increase in energy! Highway speed is way more dangerous.

Double the speed and the KE increases by four times. Very important to know

A 1 kg meteorite strikes the Moon at 11 km/s. How much KE is that?

KE = ½ × 1 kg × (11,000 m/s) 2

KE = 60,500,000 J

KE = 60.5 MJ

That is 100 times the energy of a car going at highway speed.

When falling, an object's PE due to gravity converts into KE and also heat due to air resistance.

Let's drop something!

Example: We drop this 0.1 kg apple 1 m. What speed does it hit the ground with?

At 1 m above the ground it's Potential Energy is

PE = 0.1 kg × 9.8 m/s 2 × 1 m

PE = 0.98 kg m 2 /s 2

Ignoring air resistance (which is small for this little drop anyway) that PE gets converted into KE:

Swap sides and rearrange:

½ m v 2 = KE

v 2 = 2 × KE / m

v = √( 2 × KE / m )

Now put PE into KE and we get:

v = √( 2 × 0.98 kg m 2 /s 2 / 0.1 kg )

v = √( 19.6 m 2 /s 2 )

v = 4.427... m/s

Note: for velocity we can combine the formulas like this:

The mass does not matter! It is all about height and gravity. For our earlier example:

v = √( 2gh )

v = √( 2 × 9.8 m/s 2 × 1 m )

Kinetic energy solving examples

This video gives an explanation of kinetic and contains several examples for calculating kinetic energy, mass and velocity using the kinetic

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Kinetic Energy Practice Problems

What Is Kinetic Energy? Kinetic Energy Examples

Kinetic energy is the energy possessed by a body due to its motion. The kinetic energy formula is written in terms of mass and velocity.

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Potential and Kinetic Energy

Kinetic energy is the measure of the work that an object does by virtue of its motion. Simple activities like walking, jumping, throwing

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How to Calculate Kinetic Energy: 9 Steps (with Pictures)

1. Kinetic energy depends on the velocity of the object squared. This means, when th velocity of the object is doubled, its kinetic energy

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Kinetic Energy Problemset

SHOW ALL WORK!

1. What is the kinetic energy of a jogger with a mass of 65.0 kg traveling at a speed of 2.5 m/s?

2. What is the mass of a baseball that has a kinetic energy of 100 J and is traveling at 5 m/s?

Write down what you know:

sample problem solving in kinetic energy

3. What is the kinetic energy of a 0.5 kg soccer ball that is traveling at a speed of 3 m/s?

4 What is the kinetic energy of a 1 kg pie travelling at a speed of 4 m/s ?

5. What is the kinetic energy of the pie if it is thrown at 10 m/s?

6. A student is hit with a 1 kg pumpkin pie. The kinetic energy of the pie 32 J. What was the speed of the pie?

GPE = mgh | g = 9.8 m/s 2

1. Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground.

3. A marble is on a table 2.4 m above the ground. What is the mass of the marble if it has a GPE of 568 J.

4. A box with a mass of 12.5 kg sits on the floor. How high would you need to lift it for it to have a GPE of 355J ?

5. A cart at the top of a 300 m hill has a mass of 40 kg. What is the cart’s gravitational potential energy?

6. Examine the graphic below.

What is the gravitational potential energy of the 6 kg cart as it sits the the top of the incline? _______________

What is the KINETIC ENERGY of the cart if it is moving at a speed of 2 m/s at the bottom of the ramp? _____________

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7.3: Electric Potential and Potential Difference

Learning Objectives

By the end of this section, you will be able to:

Recall that earlier we defined electric field to be a quantity independent of the test charge in a given system, which would nonetheless allow us to calculate the force that would result on an arbitrary test charge. (The default assumption in the absence of other information is that the test charge is positive.) We briefly defined a field for gravity, but gravity is always attractive, whereas the electric force can be either attractive or repulsive. Therefore, although potential energy is perfectly adequate in a gravitational system, it is convenient to define a quantity that allows us to calculate the work on a charge independent of the magnitude of the charge. Calculating the work directly may be difficult, since \(W = \vec{F} \cdot \vec{d}\) and the direction and magnitude of \(\vec{F}\) can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But we do know that because \(\vec{F}\), the work, and hence \(\Delta U\) is proportional to the test charge \(q\). To have a physical quantity that is independent of test charge, we define electric potential \(V\) (or simply potential, since electric is understood) to be the potential energy per unit charge:

Electric Potential

The electric potential energy per unit charge is

\[V = \dfrac{U}{q}. \label{eq-1}\]

Since U is proportional to q , the dependence on q cancels. Thus, V does not depend on q . The change in potential energy \(\Delta U\) is crucial, so we are concerned with the difference in potential or potential difference \(\Delta V\) between two points, where

Electric Potential Difference

The electric potential difference between points A and B , \(V_B - V_A\) is defined to be the change in potential energy of a charge q moved from A to B , divided by the charge. Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta .

\[1 \, V = 1 \, J/C \label{eq0}\]

The familiar term voltage is the common name for electric potential difference. Keep in mind that whenever a voltage is quoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, and its voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary. This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecture hall floor. It is worthwhile to emphasize the distinction between potential difference and electrical potential energy.

Potential Difference and Electrical Potential Energy

The relationship between potential difference (or voltage) and electrical potential energy is given by

\[\Delta V = \dfrac{\Delta U}{q} \label{eq1}\]

\[ \Delta U = q \Delta V. \label{eq2}\]

Voltage is not the same as energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a car battery can both have the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much more energy than the other because \(\Delta U = q\Delta V\). The car battery can move more charge than the motorcycle battery, although both are 12-V batteries.

Example \(\PageIndex{1}\): Calculating Energy

You have a 12.0-V motorcycle battery that can move 5000 C of charge, and a 12.0-V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)

To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to \(\Delta U = q\Delta V\). To find the energy output, we multiply the charge moved by the potential difference.

For the motorcycle battery, \(q = 5000 \, C\) and \(\Delta V = 12.0 \, V\). The total energy delivered by the motorcycle battery is

\[\Delta U_{cycle} = (5000 \, C)(12.0 \, V) = (5000 \, C)(12.0 \, J/C) = 6.00 \times 10^4 \, J. \nonumber\]

\[\Delta U_{car} = (60,000 \, C)(12.0 \, V) = 7.20 \times 10^5 \, J. \nonumber\]

Significance

Voltage and energy are related, but they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. A car battery has a much larger engine to start than a motorcycle. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a depleted car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.

Exercise \(\PageIndex{1}\)

How much energy does a 1.5-V AAA battery have that can move 100 C?

\(\Delta U = q\Delta V = (100 \, C)(1.5 \, V) = 150 \, J\)

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals ( A ) through whatever circuitry is involved and attract them to their positive terminals ( B ), as shown in Figure \(\PageIndex{1}\). The change in potential is \(\Delta V = V_B - V_A = +12 \, V\) and the charge q is negative, so that \(\Delta U = q \Delta V\) is negative, meaning the potential energy of the battery has decreased when q has moved from A to B .

The figure shows a headlight connected to terminals of a 12V battery. The charge q flows out from terminal A of the battery and back into terminal B of the battery.

Example \(\PageIndex{2}\): How Many Electrons Move through a Headlight Each Second?

When a 12.0-V car battery powers a single 30.0-W headlight, how many electrons pass through it each second?

To find the number of electrons, we must first find the charge that moves in 1.00 s. The charge moved is related to voltage and energy through the equations \(\Delta U = q \Delta V\). A 30.0-W lamp uses 30.0 joules per second. Since the battery loses energy, we have \(\Delta U = - 30 \, J\) and, since the electrons are going from the negative terminal to the positive, we see that \(\Delta V = +12.0 \, V\).

To find the charge q moved, we solve the equation \(\Delta U = q\Delta V\):

\[q = \dfrac{\Delta U}{\Delta V}.\]

Entering the values for \(\Delta U\) and \(\Delta V\), we get

\[q = \dfrac{-30.0 \, J}{+12.0 \, V} = \dfrac{-30.0 \, J}{+12.0 \, J/C} = -2.50 \, C.\]

The number of electrons \(n_e\) is the total charge divided by the charge per electron. That is,

\[n_e = \dfrac{-2.50 \, C}{-1.60 \times 10^{-19} C/e^-} = 1.56 \times 10^{19} \, electrons.\]

This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.

Exercise \(\PageIndex{2}\)

How many electrons would go through a 24.0-W lamp?

\(-2.00 \, C, \, n_e = 1.25 \times 10^{19} \, electrons\)

The Electron-Volt

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.

Figure \(\PageIndex{2}\) shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy that is later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by \(\Delta U = q\Delta V\), we can think of the joule as a coulomb-volt.

Part a shows an electron gun with two metal plates and an electron between the plates. The metal plates are connected to terminals of a battery and have opposite charges with a potential difference V subscript AB. Part b shows the photo of an electron gun.

The Electron-Volt Unit

On the submicroscopic scale, it is more convenient to define an energy unit called the electron-volt (eV), which is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,

\[1 \, eV = (1.60 \times 10^{-19} C)(1 \, V) = (1.60 \times 10^{-19} C)(1 \, J/C) = 1.60 \times 10^{-19} \, J.\]

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V gains 50 eV. A potential difference of 100,000 V (100 kV) gives an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V gains 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and convenient energy unit in such circumstances.

The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron-volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it acquires an energy of 30 keV (30,000 eV) and can break up as many as 6000 of these molecules \((30,000 \, eV \, : \, 5 \, eV \, per \, molecule = 6000 \, molecules)\). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can thus produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, \(K + U = constant\). A loss of U for a charged particle becomes an increase in its K . Conservation of energy is stated in equation form as

\[K + U = constant\] or \[K_i + U_i = K_f + U_f\]

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Example \(\PageIndex{3}\): Electrical Potential Energy Converted into Kinetic Energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be

\(K_i = 0\), \(K_f = \frac{1}{2}mv^2\), \(U_i = qV\), \(U_f = 0\).

Conservation of energy states that

\[K_i + U_i = K_f + U_f.\]

Entering the forms identified above, we obtain

\[qV = \dfrac{mv^2}{2}.\]

We solve this for v :

\[v = \sqrt{\dfrac{2qV}{m}}.\]

Entering values for q , V , and m gives

\[v = \sqrt{\dfrac{2(-1.60 \times 10^{-19}C)(-100 \, J/C)}{9.11 \times 10^{-31} kg}} = 5.93 \times 10^6 \, m/s.\]

Note that both the charge and the initial voltage are negative, as in Figure \(\PageIndex{2}\). From the discussion of electric charge and electric field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. These higher voltages produce electron speeds so great that effects from special relativity must be taken into account and will be discussed elsewhere. That is why we consider a low voltage (accurately) in this example.

Exercise \(\PageIndex{3}\)

How would this example change with a positron? A positron is identical to an electron except the charge is positive.

It would be going in the opposite direction, with no effect on the calculations as presented.

Voltage and Electric Field

So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between voltage and electric field. We will start with the general case for a non-uniform \(\vec{E}\) field. Recall that our general formula for the potential energy of a test charge q at point P relative to reference point R is

\[U_p = - \int_R^p \vec{F} \cdot d\vec{l}.\]

When we substitute in the definition of electric field \((\vec{E} = \vec{F}/q)\), this becomes

\[U_p = -q \int_R^p \vec{E} \cdot d\vec{l}.\]

Applying our definition of potential \((V = U/q)\) to this potential energy, we find that, in general,

\[V_p = - \int_R^p \vec{E} \cdot d\vec{l}.\]

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we can pick the integral path that is most convenient.

Consider the special case of a positive point charge q at the origin. To calculate the potential caused by q at a distance r from the origin relative to a reference of 0 at infinity (recall that we did the same for potential energy), let \(P = r\) and \(R = \infty\), with \(d\vec{l} = d\vec{r} = \hat{r}dr\) and use \(\vec{E} = \frac{kq}{r^2} \hat{r}\). When we evaluate the integral

\[V_p = - \int_R^p \vec{E} \cdot d\vec{l}\] for this system, we have

\[V_r = - \int_{\infty}^r \dfrac{kq}{r^2} dr = \dfrac{kq}{r} - \dfrac{kq}{\infty} = \dfrac{kq}{r}.\]

This result,

\[V_r = \dfrac{kq}{r}\]

is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electric field \(\vec{E}\) is produced by placing a potential difference (or voltage) \(\Delta V\) across two parallel metal plates, labeled A and B (Figure \(\PageIndex{3}\)). Examining this situation will tell us what voltage is needed to produce a certain electric field strength. It will also reveal a more fundamental relationship between electric potential and electric field.

The figure shows electric field between two plates (A and B) with opposite charges. The plates are separated by distance d and have a potential difference V subscript AB. A positive charge q is located between the plates and moves from A to B.

From a physicist’s point of view, either \(\Delta V\) or \(\vec{E}\) can be used to describe any interaction between charges. However, \(\Delta V\) is a scalar quantity and has no direction, whereas \(\vec{E}\) is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field, a scalar quantity, is represented by E .) The relationship between \(\Delta V\) and \(\vec{E}\) is revealed by calculating the work done by the electric force in moving a charge from point A to point B . But, as noted earlier, arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in Figure \(\PageIndex{3}\) to move a positive charge q from A , the positive plate, higher potential, to B , the negative plate, lower potential, is

\[W = - \Delta U = - q\Delta V.\]

The potential difference between points A and B is

\[- \Delta V = - (V_B - V_A) = V_A - V_B = V_{AB}.\]

Entering this into the expression for work yields

\[W = qV_{AB}.\]

Work is \(W = \vec{F} \cdot \vec{d} = Fd \, cos \, \theta\): here \(cos \, \theta = 1\), since the path is parallel to the field. Thus, \(W = Fd\). Since \(F = qE\) we see that \(W = qEd\).

Substituting this expression for work into the previous equation gives

\[qEd = qV_{AB}.\]

The charge cancels, so we obtain for the voltage between points A and B .

In uniform E-field only: \[V_{AB} = Ed\] \[E = \dfrac{V_{AB}}{d}\] where d is the distance from A to B , or the distance between the plates in Figure \(\PageIndex{3}\). Note that this equation implies that the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation among units is valid:

\[1 \, N/C = 1 \, V/m.\]

Furthermore, we may extend this to the integral form. Substituting Equation \ref{eq1} into our definition for the potential difference between points A and B , we obtain

\[V_{AB} = V_B - V_A = - \int_R^B \vec{E} \cdot d\vec{l} + \int_R^A \vec{E} \cdot d\vec{l}\]

which simplifies to

\[V_B - V_A = - \int_A^B \vec{E} \cdot d\vec{l}.\]

As a demonstration, from this we may calculate the potential difference between two points ( A and B ) equidistant from a point charge q at the origin, as shown in Figure \(\PageIndex{4}\).

The figure shows a charge q equidistant from two points, A and B.

To do this, we integrate around an arc of the circle of constant radius r between A and B , which means we let \(d\vec{l} = r\hat{\varphi}d\varphi\), while using \(\vec{E} = \frac{kq}{r^2} \hat{r}\). Thus,

\[\Delta V = V_B - V_A = - \int_A^B \vec{E} \cdot d\vec{l}.\]

for this system becomes

\[V_B - V_A = - \int_A^B \frac{kq}{r^2} \cdot r\hat{\varphi}d\varphi.\]

However, \(\hat{r} \cdot \hat{\varphi}\) and therefore

\[V_B - V_A = 0.\]

This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.

Example \(\PageIndex{4A}\): What Is the Highest Voltage Possible between Two Plates?

Dry air can support a maximum electric field strength of about \(3.0 \times 10^6 V/m\). Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

We are given the maximum electric field E between the plates and the distance d between them. We can use the equation \(V_{AB} = Ed\) to calculate the maximum voltage.

The potential difference or voltage between the plates is

\[V_{AB} = Ed.\]

Entering the given values for E and d gives

\[V_{AB} = (3.0 \times 10^6 V/m)(0.025 \, m) = 7.5 \times 10^4 \, V\] or \[V_{AB} = 75 \, kV.\]

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up with static electricity on dry days (Figure \(\PageIndex{5}\)).

The first photo shows a spark chamber and the second photo shows its operation.

Example \(\PageIndex{1B}\): Field and Force inside an Electron Gun

An electron gun (Figure \(\PageIndex{2}\)) has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. (a) What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a \(0.500-\mu C\) charge that gets between the plates?

Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression \(E = \frac{V_{AB}}{d}\). Once we know the electric field strength, we can find the force on a charge by using \(\vec{F} = q\vec{E}\). Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, \(F = qE\).

a. The expression for the magnitude of the electric field between two uniform metal plates is

\[E = \dfrac{V_{AB}}{d}.\] Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for \(V_{AB}\) and the plate separation of 0.0400 m, we obtain \[E = \frac{25.0 \, kV}{0.0400 \, m} = 6.25 \times 10^5 \, V/m.\]

b. The magnitude of the force on a charge in an electric field is obtained from the equation \[F = qE.\] Substituting known values gives

\[F = (0.500 \times 10^{-6}C)(6.25 \times 10^5 V/m) = 0.313 \, N.\]

Significance Note that the units are newtons, since \(1 \, V/m = 1 \, N/C\). Because the electric field is uniform between the plates, the force on the charge is the same no matter where the charge is located between the plates.

Example \(\PageIndex{4C}\): Calculating Potential of a Point Charge

Given a point charge \(q = +2.0-n C\) at the origin, calculate the potential difference between point \(P_1\) a distance \(a = 4.0 \, cm\) from q , and \(P_2\) a distance \(b = 12.0 \, cm\) from q , where the two points have an angle of \(\varphi = 24^o\) between them (Figure \(\PageIndex{6}\)).

The figure shows two points P subscript 1 and P subscript 2 at distances a and b from the origin and having an angle phi between them.

Strategy Do this in two steps. The first step is to use \(V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}\) and let \(A = a = 4.0 \, cm\) and \(B = b = 12.0 \, cm\), with \(d\vec{l} = d\vec{r} = \hat{r}dr\) and \(\vec{E} = \frac{kq}{r^2} \hat{r}.\) Then perform the integral. The second step is to integrate \(V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}\) around an arc of constant radius r , which means we let \(d\vec{l} = r\vec{\varphi}d\varphi\) with limits \(0 \leq \varphi \leq 24^o\), still using \(\vec{E} = \frac{kq}{r^2}\hat{r}\).

Then add the two results together.

Solution For the first part, \(V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}\) for this system becomes \(V_b - V_a = - \int_a^b \frac{kq}{r^2}\hat{r} \cdot \hat{r}dr\) which computes to

\(\Delta V = - \int_a^b \frac{kq}{r^2}dr = kq \left[\frac{1}{a} - \frac{1}{b}\right]\)

\(= (8.99 \times 10^9 Nm^2/C^2)(2.0 \times 10^{-9}C) \left[\frac{1}{0.040 \, m} - \frac{1}{0.12 \, m}\right] = 300 \, V\).

For the second step, \(V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}\) becomes \(\Delta V = - \int_{0^o}^{24^o} \frac{kq}{r^2} \hat{r} \cdot r\hat{\varphi}d\varphi\), but \(\hat{r} \cdot \hat{\varphi} = 0\) and therefore \(\Delta V = 0\). Adding the two parts together, we get 300 V.

We have demonstrated the use of the integral form of the potential difference to obtain a numerical result. Notice that, in this particular system, we could have also used the formula for the potential due to a point charge at the two points and simply taken the difference.

Exercise \(\PageIndex{4}\)

From the examples, how does the energy of a lightning strike vary with the height of the clouds from the ground? Consider the cloud-ground system to be two parallel plates.

Given a fixed maximum electric field strength, the potential at which a strike occurs increases with increasing height above the ground. Hence, each electron will carry more energy. Determining if there is an effect on the total number of electrons lies in the future.

Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.

Problem-Solving Strategy: Electrostatics

Practice Problems on Potential Energy

In daily use, the potential word is used a lot for things or persons which show promise inside them. “Potential” shows the possibility for action. It gives an idea of stored energy that can be converted. This is the idea behind potential energy. This concept is an integral part of mechanics and allows us to theoretically measure the energy stored inside an object. Potential energy can come through any force. For example – a stretched or compressed spring has potential energy. An object sitting at some height possesses potential energy due to height.

Potential Energy

Potential energy is the energy possessed by an object due to its position or configuration. This energy is said to be stored inside the object. Usually, potential energy is released by an object by motion. For example, a stretched spring, when released, starts moving towards its natural position and starts acquiring speed. Due to this speed, it acquires kinetic energy. To further understand this, let’s consider a ball of mass “m”.

This ball which was initially on the ground is taken to a height of “h”. External force in the form of gravity is acting on it. We know that work done by a force F on a displacement “s” on the object is given by,

In this case, force is the force of gravity and displacement from the ground to the height “h”.

F = mg, s = h

Then, work done by gravity on the object is,

Potential energy is defined as the negative of this work. Denoting the potential energy V(h).

If the ball is dropped, in that case, the potential energy decreases, and the speed increases. This means the potential energy of the object is getting converted to kinetic energy. Let’s say the speed of the ball just before touching the ground is “v”.

Potential Energy in a Spring

When the spring is kept normally, it is said to have 0 energy and is said to be in the equilibrium state. When it is stretched or compressed, and there is a certain displacement, say x, it will have certain potential energy saved in it which is given as,

P.E.= 1/2 (Kx 2 )

Where, K = Spring constant

x = displacement due to compression or expansion.

Let’s see some problems based on these concepts.

Sample Problems

Question 1: A mass of 2Kg is taken from the ground to the height of 10m. Find the potential energy of the object.

The potential energy of a mass ‘m’ at the height ‘h’ is given by, P = mgh Given: m = 2kg and g = 10 m/s 2 and h = 10m. Aim: Find the potential energy. Plugging in the values in the formula. P = mgh ⇒ P = (2)(10)(10) ⇒P = 200J Thus, the potential energy of the object is 200J.

Question 2: A mass of 5Kg is taken from the ground to the height of 100m. Find the potential energy of the object.

The potential energy of a mass ‘m’ at the height ‘h’ is given by, P = mgh Given: m = 5kg and g = 10 m/s 2 and h = 100m. Aim: Find the potential energy. Plugging in the values in the formula. P = mgh ⇒ P = (5)(10)(100) ⇒P = 5000J Thus, the potential energy of the object is 5000J.

Question 3: A mass of 5Kg is taken from the ground for 5 m uphill on the wedge. The wedge makes an angle of 30° with the ground. Find the potential energy of the block.

The potential energy of a mass ‘m’ at the height ‘h’ is given by, P = mgh This wedge is in the form of a right-angled triangle. Figure Let’s say, h is the vertical height at which the box reaches, let the slanted length be L L = 5 m Given: m = 5kg and g = 10 m/s 2 and h = 2.5m. Aim: Find the potential energy. Plugging in the values in the formula. P = mgh ⇒ P = (5)(10)(2.5) ⇒P = 125J Thus, the potential energy of the object is 125J.

Question 4: A mass of 10Kg is taken from the ground for 10m uphill on the wedge. The wedge makes an angle of 30° with the ground. Find the potential energy of the block.

The potential energy of a mass ‘m’ at the height ‘h’ is given by, P = mgh This wedge is in the form of a right-angled triangle. Figure Let’s say, h is the vertical height at which the box reaches, let the slanted length be L L = 10m Given: m = 5kg and g = 10 m/s 2 and h = 5 m. Aim: Find the potential energy. Plugging in the values in the formula. P = mgh ⇒ P = (5)(10)(5) ⇒P = 250J Thus, the potential energy of the object is 250J.

Question 5: Find the kinetic energy of the ball just before hitting the ground. Assume that initially, the ball was at a height of 10m, and its mass was 2Kg.

Answer:

Initially, at the height of 10m, the ball possesses potential energy. When it is dropped, it starts going towards the ground and its height starts decreasing. With decreasing height, velocity increases, and it acquires kinetic energy. Potential Energy at t = 0 Potential energy will be given by, P = mgh m = 2Kg, h = 10m and g = 10 m/s 2 P = mgh ⇒ P = (2)(10)(10) ⇒P = 200J When the ball is about to hit the ground, it’s potential energy has become zero and all the energy is converted into kinetic energy. Thus, K.E = 200J

Question 6: Find the velocity of the ball just before hitting the ground. Assume that initially, the ball was at a height of 100m, and its mass was 4Kg.

Initially, at the height of 10m, the ball possesses potential energy. When it is dropped, it starts going towards the ground and its height starts decreasing. With decreasing height, velocity increases, and it acquires kinetic energy. Potential Energy at t = 0 Potential energy will be given by, P = mgh m = 4Kg, h = 100m and g = 10 m/s 2 P = mgh ⇒ P = (4)(100)(10) ⇒P = 4000J When the ball is about to hit the ground, its potential energy has become zero and all the energy is converted into kinetic energy. Thus, K.E = 4000J The formula for K.E is, K.E = m = 4Kg and v = ?. Plugging the values in the formula K.E = ⇒ 4000 = ⇒2000 = v 2 ⇒ v = 10√20 m/s ⇒ v = 20√5 m/s

Question 7: The entire potential energy of a ball is transformed into its kinetic energy by coming on the ground from a certain height. The height at which the ball was initially placed was 10m. The mass of the ball is 1 kg. Find the gain in the kinetic energy.

Since, the all the potential energy present in the ball is transferred into its kinetic energy, Potential energy of the ball = Final gain in the kinetic energy P= mgh m= 1kg, h= 10m, g= 9.8m/sec 2 P= 1× 10× 9.8 P= 98 Joule Therefore, the final gain in kinetic energy is 98 Joules. K.E= 98 Joule.

Question 8: Explain the existence of potential energy,

a. Due to its position

b. Due to the state in which the object is.

Potential energy can actually be present in two different cases, a. Due to its position Suppose an object is stable on the ground, now by applying some energy, it is taken at a certain height. The object present at a certain height will have energy saved in the form of potential energy. It is given as, P.E= mgh b. Due to the state in which object is present. When a spring is kept in normal state, it is said to have 0 energy, but when the same spring is either compressed or stretched, it obtains potential energy which is given as, P.E. = 1/2 (Kx 2 ) Where, x= displacement, K= Spring constant.

Question 9: A spring is stretched upto 9cm, the spring constant of the spring is 2 N/m. Find the value of Potential energy stored in the spring?

The potential energy stored in a spring is given as, P.E= 1/2 (Kx 2 ) K= 2 N/m, x= 9cm= 0.09m P.E= 1/2 (2× (0.09) 2 ) P.E= 8.1 × 10 -3 Joules.

Question 10: Two objects are kept at different heights, first object is at 5meters and second object is placed at 15 meters. Which object has more Potential energy if the first object is 5 times heavier than the second?

Object 1: Height= 5 m, Mass= 5 m P.E 1 = 5 m× 5× 9.8 P.E 1 = 245m Joules. Object 2: Height= 15 m, Mass= m P.E 2 = 15× m× 9.8 P.E 2 = 147m Joules. Therefore, even when the first object is kept at lesser height, due to its weight, the first object has more potential energy.

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Mechanical Energy Problem Solutions

Home » Unit 5: Work, Power, Mechanical Energy, and Simple Machines » Mechanical Energy » Mechanical Energy Problem Solutions

Mechanical Energy Problems and Solutions

See examples of mechanical energy problems involving kinetic energy, potential energy, and the conservation of energy. Check your work with ours.

1. How much gravitational potential energy do you have when you lift a 15 N object 10 meters off the ground?

2. How much gravitational potential energy is in a 20 kg mass when 0.6 meters above the ground?

3. How much gravitational potential energy does a 35 kg boulder have when 30 meters off the ground?

4. How many times greater is an objects potential energy when three times higher?

If you need help on ratio problems click the link below:

Rule of Ones: analyzing equations to determine how other variables change

5. How much kinetic energy does a 0.15 kg ball thrown at 24 m/s have?

6. How many times greater is the kinetic energy of a ball that is going five times faster?

7. How much kinetic energy does a 1.2 kg ball have the moment it hits the ground 3.5 meters below when it starts from rest?

I cancelled out the initial kinetic energy because:

I cancelled out the final potential energy because:

8. How fast is a 1.2 kg ball traveling the moment it hits the ground 3.5 meters below when it starts from rest?

(Note: In many of these problems I could cancel out mass but did not since it was provided)

Since I did not cancel out mass I could answer the following questions if asked:

If I cancelled out mass in my work it would not show the actual initial potential energy since PE i = mgh and not just gh.

9. A 3.5 kg ball fell from a height of 12 meters. How fast is it traveling when its still 5 meters off the ground?

10. An 85kg roller coaster cart is traveling 4 m/s at the top of a hill 50 meters off the ground. How fast is it traveling at top of a second hill 20 meters off the ground?

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7. How much kinetic energy does a 1.2 kg ball have the moment it hits the ground 3.5 meters below when it starts from rest? I cancelled out the initial kinetic energy because: KE i = ½ mv f2 KE i = (½) (3.5) (0 2) = 0 J I cancelled out the final potential energy because: PE f = mgh f PE f = (3.5) (9.8) (0) = 0 J 8.
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