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## Potential And Kinetic Energy Example Problem – Work and Energy Examples

## Potential And Kinetic Energy Example Problem

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

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

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

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

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

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 )

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

The maximum height of the cart is 10.2 m.

## Related Posts

## Kinetic Energy Formula | Problems (With Solutions)

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

## Kinetic Energy Practice Problems

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

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

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

- Thermal Energy Equation
- Potential Energy Formula
- Gravitational Potential Energy Formula
- Electric Potential Energy Formula
- Elastic Potential Energy Formula
- Rotational Kinetic Energy Formula
- Electrical Energy Equation
- Mechanical Energy Formula
- Photon Energy Equation
- Conservation of Energy Formula
- Types of Energy
- Kinetic Energy Examples

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## How to Calculate Kinetic Energy

Last Updated: September 22, 2022 References

## Understanding Kinetic Energy

- Tare the balance. Before you weigh your object, you must set it to zero. Zeroing out the scale is called taring. [6] X Research source
- Place your object in the balance. Gently, place the object on the balance and record its mass in kilograms.
- If necessary, convert grams to kilograms. For the final calculation, the mass must be in kilograms.

- Velocity is defined by the equation, displacement divided by time: V = d/t . Velocity is a vector quantity, meaning it has both a magnitude and a direction. Magnitude is the number value that quantifies the speed, while the direction is the direction in which the speed takes place during motion.
- For example, an object’s velocity can be 80 m/s or -80 m/s depending on the direction of travel.
- To calculate velocity, simply divide the distance the object traveled by the time it took to travel that distance.

## Calculating Kinetic Energy

## Using Kinetic Energy to Find Velocity or Mass

- Multiply mass by 0.5: 0.5 x 30 = 15
- Divide kinetic energy by the product: 500/15 = 33.33
- Square root to find velocity: 5.77 m/s
- Square the velocity: 5 2 = 25
- Multiply by 0.5: 0.5 x 25 = 12.5
- Divide kinetic energy by product: 100/12.5 = 8 kg

## Community Q&A

## Video . By using this service, some information may be shared with YouTube.

## You Might Also Like

- ↑ https://byjus.com/physics/potential-energy/
- ↑ https://www.khanacademy.org/science/physics/work-and-energy/work-and-energy-tutorial/a/what-is-kinetic-energy
- ↑ https://sciencing.com/sources-of-kinetic-energy-12298540.html
- ↑ https://study.com/academy/lesson/tare-weight-vs-net-weight.html
- ↑ http://www.physicsclassroom.com/class/1DKin/Lesson-1/Speed-and-Velocity
- ↑ http://www.physicsclassroom.com/class/energy/u5l1c.cfm

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## Kinetic energy problems

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.

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

## Tricky kinetic energy problems

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

Think carefully and try to solve this problem yourself.

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## Kinetic energy problem solving examples

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## Kinetic Energy Examples

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## Kinetic Energy Examples

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

## Kinetic Energy

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

- A car with a mass of 700 kg is moving with a speed of 20m/s. Calculate the kinetic energy of the car.
- A cyclist and bike have a total mass of 100 kg and a speed of 15 m/s. Calculate the kinetic energy.
- A tennis ball is traveling at 50 m/s and has a kinetic energy of 75 J. Calculate the mass of the tennis ball.

Kinetic Energy - IGCSE Physics

Calculations using the kinetic energy formula.

- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
- A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
- A 5000kg truck has 400000J of kinetic energy. How fast is it moving?
- A car traveling at 10 m/s has 200000J of kinetic energy. What is the mass of the car?

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## Rotational Kinetic Energy - Problem Solving

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## 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 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, \[\begin{array}{l} g(h - R) = \frac{7}{{10}}{v^2}\\ v = \sqrt {\frac{{10g(h - R)}}{7}} \end{array}\]

## Kinetic energy problem solving examples

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## Potential and Kinetic Energy

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## Kinetic Energy Practice Problems

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## Examples of Kinetic Energy Problems.

## Practice Problems on Kinetic Energy

## Kinetic Energy Formula

the velocity with which the body is travelling is v .

The Kinetic energy is articulated in Kgm 2 /s 2

## Kinetic Energy Solved Examples

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

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

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

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## Register with BYJU'S & Download Free PDFs

## Potential and Kinetic Energy

Energy is the capacity to do work .

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

## Potential Energy and Kinetic Energy

- when raised up has potential energy (the energy of position or state)
- when falling down has kinetic energy (the energy of motion)

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

- a raised hammer has PE due to gravity.
- fuel and explosives have Chemical PE
- a coiled spring or a drawn bow also have PE due to their 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:

- m is the objects mass (kg)
- g is the "gravitational field strength" of 9.8 m/s 2 near the Earth's surface
- h is height (m)

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

## 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 highway speed of 28 m/s (about 100 km/h or 60 mph)?

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

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.

## 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

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

Now put PE into KE and we get:

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

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:

- Energy is the ability to do work
- Potential Energy (PE) is stored energy due to position or state
- Kinetic Energy (KE) is energy of motion

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

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## Kinetic Energy Problemset

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?

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?

## GPE = mgh | g = 9.8 m/s 2

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## 7.3: Electric Potential and Potential Difference

## Learning Objectives

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

- Define electric potential, voltage, and potential difference
- Define the electron-volt
- Calculate electric potential and potential difference from potential energy and electric field
- Describe systems in which the electron-volt is a useful unit
- Apply conservation of energy to electric systems

## Electric Potential

The electric potential energy per unit charge is

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

## Electric Potential Difference

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

## Potential Difference and Electrical Potential Energy

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

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

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

Similarly, for the car battery, \(q = 60,000 \, C\) and

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

## Exercise \(\PageIndex{1}\)

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

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

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

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,

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

## 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 Electron-Volt Unit

## Conservation of Energy

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

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

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

Conservation of energy states that

Entering the forms identified above, we obtain

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

Entering values for q , V , and m gives

## Exercise \(\PageIndex{3}\)

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

## Voltage and Electric Field

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

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

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

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

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

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

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

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

Substituting this expression for work into the previous equation gives

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

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

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

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

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

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

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

The potential difference or voltage between the plates is

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.)

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

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

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

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

Then add the two results together.

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

## Exercise \(\PageIndex{4}\)

## Problem-Solving Strategy: Electrostatics

- Examine the situation to determine if static electricity is involved; this may concern separated stationary charges, the forces among them, and the electric fields they create.
- Identify the system of interest. This includes noting the number, locations, and types of charges involved.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. Determine whether the Coulomb force is to be considered directly—if so, it may be useful to draw a free-body diagram, using electric field lines.
- Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is important to distinguish the Coulomb force F from the electric field E , for example.
- Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines as requested.
- Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involved reasonable?

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## Potential Energy

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

x = displacement due to compression or expansion.

Let’s see some problems based on these concepts.

## Sample Problems

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.

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.

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.

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.

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

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

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,

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.

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.

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

## Mechanical Energy Problems and Solutions

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?

I cancelled out the initial kinetic energy because:

I cancelled out the final potential energy because:

(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:

- How much mechanical energy did you have at the beginning? (41.6 J)
- How much kinetic energy did you have at the beginning? (0 J)
- How much potential energy did you have at the beginning? (41.6 J)
- How much potential energy do you have at the end? (0 J)

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## COMMENTS

Using the kinetic energy equation. A cannon launches a 3.0\,\text {kg} 3.0kg pumpkin with 110\,\text J 110J of kinetic energy.

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.

Kinetic energy is a scalar quantity, and it is expressed in Joules. Sample Problems Question 1: A ball has a mass of 2Kg, suppose it travels at 10m/s. Find the kinetic energy possessed by it. Answer: Given: m = 2Kg, and v = 10m/s The KE is given by, K.E = K.E = ⇒ K.E = ⇒ K.E = 100J

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

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

The first step to solving this problem is to plug in all of the variables that are known. Example 1: What is the velocity of an object with a mass of 30 kg and a kinetic energy of 500 J? KE = 0.5 x mv2 500 J = 0.5 x 30 x v2 Example 2: What is the mass of an object with a kinetic energy of 100 J and a velocity of 5 m/s? KE = 0.5 x mv2

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 = ½ mv2 KE = ? m = 150kg v = 15m/s KE = ½ (150kg) (15 m/s)2 KE = ½ (150kg)(225) KE = 16875J 2. An object has a kinetic energy of 25 J and a mass of 34 kg , how fast is the object moving?

Solution: 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. K = m (3v) 2 2 K = m 9v 2 2 K = 9 mv 2 2

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. ... 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.

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

Kinetic energy is the energy stored in moving objects. Stationary objects have no kinetic energy. E k = 0.5 × m × v 2 Examples: A car with a mass of 700 kg is moving with a speed of 20m/s. Calculate the kinetic energy of the car. A cyclist and bike have a total mass of 100 kg and a speed of 15 m/s. Calculate the kinetic energy.

Steps for Solving Kinetic Energy Problems Step 1: List the given mass and velocity of the object. Step 2: If necessary, convert the mass and velocity values so they have units of kilograms and...

Rotational Kinetic Energy - Problem Solving. 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 ...

Video transcript. I'm going to show you some examples of how to solve problems involving work. Imagine a 4 kilogram trashcan. The trashcan is disgusting. So someone ties a string to it and pulls on the string with a force of 50 newtons. The force of kinetic friction on the trashcan while it slides is 30 newtons.

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:.

Solving Kinetic Energy Sample Problem with Mrs. Aki - YouTube Use this video to help you solve problems involving kinetic energy. Use this video to help you solve problems involving...

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 k.

Kinetic Energy, mass and speed. (You need to be able to use the equation ! E k = 1 2 mv2) • carry out calculations involving energy, work, power and the principle of conservation of energy. (You can be asked to solve problems similar to the lift example we looked at in class) The examples in this handout are designed to help prepare you for

Kinetic Energy The formula is: KE = ½ m v 2 Where m is the object's mass (kg) v is the object's speed (m/s) 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 m v 2 KE = ½ × 1500 kg × (14 m/s) 2 KE = 147,000 kg m 2 /s 2 KE = 147 kJ Let's double the speed!

Kinetic and potential energy practice problems - This Kinetic and potential energy practice problems supplies step-by-step instructions for solving all math. Math Index ... Math is a way of solving problems by using numbers and equations.

Any object that is moving has kinetic energy. An example is a baseball that has been thrown. The kinetic energy depends on both mass and velocity and can be expressed mathematically as follows: ... Calculating: Have students practice problems solving for potential energy and kinetic energy: If a mass that weighs 8 kg is held at a height of 10 m ...

Kinetic energy solving examples - This video gives an explanation of kinetic and contains several examples for calculating kinetic energy, mass and velocity ... Kinetic Energy Practice 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

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. m =. g =. Answer: h =. GPE =. 2. An apple in a tree has a gravitational potential energy of 175 J and a mass of 0.36 g.

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 ...

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. Answer: 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

Steps for Solving Conservation of Energy Problems Step 1: Make a list of all known quantities given in the problem such as the object's mass, its initial and final height, and its initial...

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.

Solving Kinetic Energy Problems. Calculate the unknown variable in the equation for kinetic energy, where kinetic energy is equal to one half times the mass multiplied by velocity squared. Decide mathematic questions. Download full answer. Explain mathematic question.