The kinetic energy of an object depends on more factors than just the gravitational potential energy of an object.

C3.1 Hummingbird

Kinetic energy is the energy of motion. Imagine a large bald eagle flying toward you or a tiny hummingbird flying toward you. Which bird has more kinetic energy? Why?

Gravitational potential energy is dependent upon the work done on an object. Which would require more work, lifting a ping-pong ball off the ground or lifting a bowling ball? Why?

By the end of this lesson, you will know how to calculate gravitational potential energy and kinetic energy.

By the end of this lesson, you will be able to

relate gravitational potential energy to work done using «math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «/math».
calculate kinetic energy using «math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «mo»=«/mo» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/math»
understand scientific notation and significant digits

Significant Digits

https://adlc.wistia.com/medias/p8ku1god3z

This video reviews the rules of significant digits in scientific calculations. When performing calculations and collecting data, the accuracy of the values being used is very important! This will help get you in the right mindset for this lesson.

Gravitational potential energy can be found in many natural and technological systems.

C3.2 Logging crane

When work is done against gravity, the force is the weight of the object or the force of gravity working on it. The distance over which work is done against gravity must be the height that you lift the object.

How is the amount of gravitational potential energy in a system calculated?

Gravitational potential energy is determined by the mass of the object, acceleration due to gravity, and the height the object is being lifted against the force of gravity.

E_{p} = m g h E_{p} = g r a v i t a t i o n a l p o t e n t i a l e n e r g y (J) m = m a s s (k g) g = a c c e l e r a t i o n d u e t o g r a v i t y = 9.81 m / s^{2} h = h e i g h t (m)

Remember that work is also a measurement of energy, so its units are joule (J) as well.

Examples

Example 1

A picture with a mass of 3.75 kg hangs on a wall 2.3 m above the floor. What is the gravitational potential energy of the picture relative to the floor?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

$E_{p} = ? m = 3.75 kg g = 9.81 \frac{m}{s^{2}} h = 2.3 m$

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

Step 3: Substitute the values into the formula.

$E_{p} = (3.75 kg) (9.81 \frac{m}{s^{2}}) (2.3 m)$

Step 4: Calculate the answer.

E_p = 84.611 25 J

The answer must be rounded to two significant digits.

The gravitational potential energy of the picture is 85 J.

Example 2

A roller coaster raises its passengers a vertical distance of 3 650 cm above the ground. The total mass of the roller coaster and passengers is 6.23 × 10³ kg.

What is the gravitational potential energy of the roller coaster and its passengers relative to the ground?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

$E_{p} = ? m = 6.23 \times 10^{3} kg g = 9.81 \frac{m}{s^{2}} h = 3 650 cm = ? m$

Since h is measured in metres, we have to convert our units of cm to m.

$3 650 cm \times (\frac{1 m}{100 cm}) = 36.5 m$

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

Step 3: Substitute the values into the formula.

$E_{p} = (6.23 \times 10^{3} kg) (9.81 \frac{m}{s^{2}}) (36.5 m)$

Step 4: Calculate the answer.

E_p = 2 230 744.95 J

The answer must be rounded to three significant digits.

2 230 744.95 J cannot be rounded to three significant digits, so it must be put into scientific notation.
Move the decimal point to the left until your answer is between 1 and 10.
2 230 744.95: Move the decimal six places to the left to become 2.230 744 95.
Six decimals to the left is indicated by a 10⁶ (“6” for moving six places to the left).
Round the value of 2.23074495 to 3 sig digs: 2.23.

The gravitational potential energy of the roller coaster and its passengers is 2.23 × 10⁶ J.
How much work did the engine do on the roller coaster and its passengers in carrying them to the top?

Since the work required to lift the roller coaster and its passengers was the force required to raise them to the height against gravity, then the gravitational potential energy and work will be the same.
- Step 1
Step 1:

W = 2.23 × 10 ⁶ J

Since the work required to lift the roller coaster and its passengers was the force required to raise them to the height against gravity, then the gravitational potential energy and work will be the same.

Example 3

How far would you have to lift a 7 . 25 kg bowling ball to give it 242 J of gravitational potential energy?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

$E_{p} = 242 J m = 7.25 kg g = 9.81 \frac{m}{s^{2}} h = ?$

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

To isolate h, you must divide each side by mg. To move m and g to the other side, you must use the opposite operation.

$\frac{E_{p}}{m g} = \frac{m g h}{m g}$

Now, cancel the like terms.

$\frac{E_{p}}{m g} = h or h = \frac{E_{p}}{m g}$

Step 3: Substitute the values into the formula.

$h = \frac{242 J}{(7.25 kg) (9.81 \frac{m}{s^{2}})}$

Step 4: Calculate the answer.

h = 3.402...m

The answer must be rounded to three significant digits.

The bowling ball would need to be lifted 3.40 m.

Example 4

A child is swinging on a swing in her backyard. She reaches a height of 2.64 m above the ground. If she had 607 J of gravitational potential energy relative to the ground at its highest point, what is the child’s mass?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

$E_{p} = 607 J m = ? g = 9.81 \frac{m}{s^{2}} h = 2.64 m$

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

To isolate m, you must divide each side by gh. To move g and h to the other side, you must use the opposite operation.

$\frac{E_{p}}{g h} = \frac{m g h}{g h}$

Now, cancel the like terms.

$\frac{E_{p}}{g h} = m or m = \frac{E_{p}}{h g}$

Step 3: Substitute the values into the formula.

$m = \frac{607 J}{(2.64 m) (9.81 \frac{m}{s^{2}})}$

Step 4: Calculate the answer.

m = 23.437...kg

The answer must be rounded to three significant digits.

The child's mass is 23.4 kg.

Please read pages 173 to 175 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how the amount of gravitational potential energy can be calculated using the E_p = mgh formula. Remember, if you have any questions or do not understand something, ask your teacher!

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

A 57.3 g tennis ball is held above the ground as the tennis player prepares to serve it to his opponent. If he holds the ball at a height of 2.45 m, what is the ball’s gravitational potential energy relative to the ground?

Check your work.

Step 1: List the variables.

E_{p} = ? m = 57.3 g = 0.0573 kg g = 9.81 \frac{m}{s^{2}} h = 2.45 m

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

Step 3: Substitute the values into the formula.

E_{p} = (0.0573 kg) (9.81 \frac{m}{s^{2}}) (2.45 m)

Step 4: Calculate the answer.

E_p = 1.377...J

The answer must be rounded to three significant digits.

The gravitational potential energy of the tennis ball is 1.38 J.

You carry a heavy box filled with books up three flights of stairs for a total vertical height of 10.9 m. If you do 1 450 J of work, what is the mass of the box filled with books?

Check your work.

Step 1: List the variables.

\begin{array}{rcl} E_{p} & = & 1 450 J \\ m & = & ? \\ g & = & 9.81 \frac{m}{s^{2}} \\ h & = & 10.9 m \end{array}

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

To isolate m, you must divide each side by gh. To move g and h to the other side, you must use the opposite operation.

\frac{E_{p}}{g h} = \frac{m g h}{g h}

Now, cancel the like terms.

\frac{E_{p}}{g h} = m or m = \frac{E_{p}}{h g}

Step 3: Substitute the values into the formula.

m = \frac{1 450 J}{(10.9 m) (9.81 \frac{m}{s^{2}})}

Step 4: Calculate the answer.

m = 13.560...kg

The answer must be rounded to three significant digits.

The mass of the box filled with books is 13.6 kg.

Potential Energy on Shelves © 2017 Explore Learning

Work through this activity to discover how gravity gives objects potential energy because of their position above the floor.

Remember that potential energy is the energy an object has because of its position or shape.

Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.

Procedure
Analysis

Click on the play button to open the activity. This activity can also be accessed in the Online Resources for Print Students section of your online course.
Which object on the “SIMULATION” pane has the least potential energy? Why?

Check your work.

The ball most likely has the least potential energy because it is sitting at ground level.

Click on the “TABLE” tab. The potential energy (E_p) of each object is given in joules (J).

List the objects in order from lowest to highest potential energies.

Check your work.

Ball, paper, clips

Activity A: Factors Affecting E_p	Get the Gizmo ready: Select the “BAR CHART” tab and turn on “Show numerical values.”	C3.3 Image of bar chart
What factors affect how much gravitational potential energy an object has?

Which of the following factors do you think affect an object’s potential energy: mass, vertical position, velocity, horizontal position?

Check your work.

mass and vertical position

Drag the ball to the 1 m shelf on the “SIMULATION” pane.
What is the ball’s potential energy (E_p)?

Check your work.

0.98 J

Move the ball to the 2 m shelf. What is its potential energy now?

Check your work.

1.96 J

What do you think the ball’s potential energy will be on the 3 m shelf?

Check your work.

2.94 J

The 4 m shelf?

Check your work.

3.92 J

Use the Gizmo to check your answers. (Click the control on the bar graph to zoom out.)
Move the ball from side to side (left to right) while trying to keep it at the same height.
How does changing the horizontal position of the ball affect its potential energy?

Check your work.

Changing the ball’s horizontal position does not affect its potential energy.

Place the ball and the paper on the same shelf.
Which object has more potential energy?

Check your work.

The ball

Why do you think their potential energies are different?

Check your work.

Your answer will be a variation of the following. The ball has more mass than the paper.

Activity B: Calculating E_p	Get the Gizmo ready: You will need a calculator to complete this activity.	C3.4 Image of various objects
Use the gravitational potential energy equation to determine the weight, mass, and potential energy of various objects. E_p = weight (W) x height

Position all three objects on the 1 m shelf and fill in the third column of the table.

Object	Height (m)	E_p (J or N•m)	Weight (N)
Ball	1
Clips	1
Paper	1

Check your work.

Object	Height (m)	E_p (J or N•m)	Weight (N)
Ball	1	0.98	0.98
Clips	1	0.39	0.39
Paper	1	0.08	0.08

For each object, substitute the values you know into the gravitational potential energy equation to solve for weight. Record each object’s weight in the fourth column.

Check your work.

For example, to calculate the weight of the ball:

E_p = weight (W) x height (h)

To isolate weight, you must divide each side by height. To move h to the other side, you need to use the opposite operation.

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Now, cancel the like terms.

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Suppose the clips were placed on the 5 m shelf. What would their gravitational potential energy be? (Show your work.)

Check your work.

E_p = Wh
E_p = (0.39 N)(5 m)
E_p = 1.95 J or N•m

Use the Gizmo to check your answer.

NOTE

The Gizmo gives a value of 1.96 J.


Activity C: Work and E_p	Get the Gizmo ready: Place the ball, clips, and paper at 0 m.	C3.5 image of various objects
How much work is done to lift the ball, clips, and paper?

How much potential energy do the ball, clips, and paper have now?

Check your work.

0.00 J
Use the weight of the ball that you calculated in activity B to determine how much work would be required to lift the ball 2 metres above the zero position.

Check your work.

W = Fd
W = (0.98 N)(2 m)
W = 1.96 J or N•m
Move the ball to the 2 m shelf.

How much potential energy does the ball have now?

Check your work.

1.96 J

How does the ball’s potential energy relate to the amount of work needed to place the ball on the 2 m shelf?

Check your work.

The two values are equal.

How much work would be needed to lift the ball from the 2 m shelf to the 5 m shelf and how much potential energy would it have on the 5 m shelf?

Check your work.

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Please return to the top of this page and click on analysis to complete the analysis questions.

What is the relationship between an object’s height above the ground and its gravitational potential energy?

Check your work.

As an object’s height increases, so does its potential energy.
What two factors affect how much gravitational potential energy an object has?

Check your work.

Height above the ground and mass.
What is the gravitational potential energy of a 4 000 kg elephant hoisted 20 m above the earth's surface?
Check your work.
Step 1

Step 2

Step 3

Step 4

Step 1: List the variables.

«math»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»=«/mo»«mo»?«/mo»«mspace linebreak=¨newline¨»«/mspace»«mi»m«/mi»«mo»=«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«mspace linebreak=¨newline¨»«/mspace»«mi»g«/mi»«mo»=«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi»h«/mi»«mo»=«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

E_p = mgh

Step 3: Substitute the values into the formula.

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Step 4: Calculate the answer.

E_p = 784 800 J

The answer must be rounded to two significant digits.

784 800 J cannot be rounded to two significant digits, so it must be put into scientific notation.
Move the decimal point to the left until your answer is between 1 and 10.
784 800: Move the decimal five places to the left to become 7.848 00.
Five decimals to the left is indicated by a «math» «msup mathcolor=¨#FFFFFF¨» «mn mathcolor=¨#FFFFFF¨»10«/mn» «mn»5«/mn» «/msup» «/math» (“«math» «mn mathcolor=¨#FFFFFF¨»5«/mn» «/math»” for moving five places to the left).
Round the value of 7.848 00 to two significant digits: 7.8.

The gravitational potential energy of the elephant is 7.8 × 10⁵ J.
Objects in motion have kinetic energy. As objects fall, their potential energy is converted into kinetic energy. How much kinetic energy do you think the ball would have just before it hit the floor if it were dropped from a 2 m shelf? Explain your answer.

Check your work.

The ball would have 1.96 J of kinetic energy because by the time the ball reached the floor, all of its potential energy would have been converted to kinetic energy.

Kinetic energy can be found in many natural and technological systems.

C3.6 Billiard’s 8 ball rack break in motion

Any object that is in motion has kinetic energy, whether it is an electron in the flow of electricity through a wire or a billiard ball rolling across the table.

How is the amount of kinetic energy in a system calculated?

Kinetic energy is determined by the mass of the object and the velocity that it is travelling at.

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Examples

Example 1

A freight elevator with a mass of 124 kg is moving with a speed of 2.4 m/s. What is the elevator’s kinetic energy?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»124«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo».«/mo»«mn»4«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

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Step 3: Substitute the values into the formula.

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Step 4: Calculate the answer.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mn»357«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/math»

The answer must be rounded to two significant digits.

357.1 J cannot be rounded to two sig digs, so it must be put into scientific notation.
Move the decimal point to the left until your answer is between 1 and 10.
357.1: Move the decimal two places to the left to become 3.571.
Two decimals to the left is indicated by a 10² (“2” for moving two places to the left).
Round the value of 3.571 to two significant digits: 3.6.

The kinetic energy of the elevator is 3.6 × 10 ² J.

Example 2

A wrecking ball has a mass of 365 kg. If it is moving toward the building it is about to demolish at a speed of 5.42 m/s, what is its kinetic energy in kilojoules?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»365«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»42«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

Step 3: Substitute the values into the formula.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»365«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»5«/mn»«mo».«/mo»«mn»42«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/math»

Step 4: Calculate the answer.

E_k = 5 361.193 J

The answer must be rounded to three significant digits and converted to kilojoules.

1 000 J = 1 kJ

«math» «mfenced» «mrow» «mn»5«/mn» «mo»§#160;«/mo» «mn»361«/mn» «mo».«/mo» «mn»193«/mn» «/mrow» «/mfenced» «mo»§#215;«/mo» «mfenced» «mfrac» «mrow» «mn»1«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/mrow» «mrow» «mn»1«/mn» «mo»§#160;«/mo» «mn»000«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «/mfrac» «/mfenced» «mo»=«/mo» «mn»5«/mn» «mo».«/mo» «mn»361«/mn» «mo»§#160;«/mo» «mn»193«/mn» «mo»§#160;«/mo» «mi»kJ«/mi» «/math»

The kinetic energy of the wrecking ball is 5.36 kJ.

Example 3

An electron is moving at a speed of 2.19 x 10⁷ m/s and has 2.18 x 10^–16 J of kinetic energy. What is the mass of the electron?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo».«/mo»«mn»18«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mrow»«mo»-«/mo»«mn»16«/mn»«mo»§#160;«/mo»«/mrow»«/msup»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»7«/mn»«/msup»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

To isolate m, you must divide each side by «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/math». To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math»and v² to the other side, you need to use the opposite operation.

«math»«mfrac»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mstyle»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mstyle»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mrow»«mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mstyle»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfrac»«/math»

Now, cancel the like terms and multiple by 2 to get rid of «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math».

«math»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mfrac»«mo»=«/mo»«mi»m«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»m«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mfrac»«/math»

Step 3: Substitute the values into the formula

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«mfenced»«mrow»«mn»2«/mn»«mo».«/mo»«mn»18«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mrow»«mo»-«/mo»«mn»16«/mn»«mo»§#160;«/mo»«/mrow»«/msup»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfenced»«/mrow»«msup»«mfenced»«mrow»«mn»2«/mn»«mo».«/mo»«mn»19«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»7«/mn»«/msup»«mstyle displaystyle=¨true¨»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mstyle»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mfrac»«/math»

Step 4: Calculate the answer.

m = 9.0907...×10 ^–31 kg

The answer must be rounded to three significant digits.

The mass of the electron is 9.09 × 10^–31 kg.

Example 4

What is the speed of a 0.148 kg billiard ball that has 13.2 J of kinetic energy?
- Step 1
- Step 2
- Step 3
- Step 4
Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»13«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»148«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

To isolate v, you must divide each side by «math»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/math». To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math» and m to the other side, you must use the opposite operation.

«math»«mfrac»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/mstyle»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mstyle»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/mstyle»«/mfrac»«/math»

Now, cancel the like terms and multiple by 2 to get rid of «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math». Then, take the square root, as that will remove the square on the v.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»o«/mi»«mi»r«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»v«/mi»«mo»=«/mo»«msqrt»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«/msqrt»«/mtd»«/mtr»«/mtable»«/math»

Step 3: Substitute the values into the formula.

«math» «mi»v«/mi» «mo»=«/mo» «msqrt» «mfrac» «mrow» «mn»2«/mn» «mfenced» «mrow» «mn»13«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «/mfenced» «/mrow» «mrow» «mn»0«/mn» «mo».«/mo» «mn»148«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfrac» «/msqrt» «/math»

Step 4: Calculate the answer.

v = 13.355...m/s

The answer must be rounded to three significant digits.

The speed of the billiard ball is 13.4 m/s.

Please read pages 179 to 181 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how the amount of kinetic energy can be calculated using the «math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math» formula. Remember, if you have any questions, or do not understand something, ask your teacher!

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

A student goes out for a jog in the morning and can travel at a speed of 1.72 m/s. If her mass is 53 000 g what is the student’s kinetic energy?

Check your work.

Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»53«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#215;«/mo»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfrac»«/mfenced»«mo»=«/mo»«mn»53«/mn»«mo».«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo».«/mo»«mn»72«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

Step 3: Substitute the values into the formula.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»53«/mn»«mo».«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»1«/mn»«mo».«/mo»«mn»72«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/math»

Step 4: Calculate the answer.

E_k = 78.3976 J

The answer must be rounded to three significant digits.

The kinetic energy of the jogger is 78.4 J.

You are paddling a canoe, and the combined mass of the canoe and your body is 117 kg. If you and the canoe have a total kinetic energy of 77 J, how fast are you paddling the canoe?

Check your work.

Step 1: List the variables.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»77«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»117«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»

Step 2: Identify the correct formula and rearrange if necessary.

«math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

To isolate v, you must divide each side by «math»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/math». To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math»and m to the other side, you must use the opposite operation.

«math»«mfrac»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/mstyle»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mstyle»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mrow»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/mstyle»«/mfrac»«/math»

Now, cancel the like terms and multiple by 2 to get rid of «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math». Then, take the square root, as that will remove the square on the v.

«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»o«/mi»«mi»r«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»v«/mi»«mo»=«/mo»«msqrt»«mfrac»«mrow»«mn»2«/mn»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mrow»«mi»m«/mi»«/mfrac»«/msqrt»«/mtd»«/mtr»«/mtable»«/math»

Step 3: Substitute the values into the formula.

«math» «mi»v«/mi» «mo»=«/mo» «msqrt» «mfrac» «mrow» «mn»2«/mn» «mfenced» «mrow» «mn»77«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «/mfenced» «/mrow» «mrow» «mn»117«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfrac» «/msqrt» «/math»

Step 4: Calculate the answer.

v = 1.147...m/s

The answer must be rounded to two significant digits.

The speed of the canoe is 1.1 m/s.

A great science teacher risks his life to explain gravitational potential and kinetic energy. This video reviews what gravitational potential and kinetic energy are and the conversion of one to other.

Calculating the gravitational potential energy and kinetic energy of an object requires an understanding of many factors.

C3.7 Roller coaster

You now have an understanding of the fact that gravitational potential energy is dependent upon an object’s mass and its relative position and movement to the surface of Earth. Kinetic energy is energy of motion and is dependent upon an object’s speed and mass.

In the next lesson, we will look at how energy is conserved when transformations occur between gravitational potential energy and kinetic energy.

This video reviews what gravitational potential and kinetic energy are and the factors that influence the amount of energy an object has.

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Site:	MoodleHUB.ca 🍁
Course:	Science 10 [5 cr] - AB Ed copy 1
Book:	Lesson 3 Gravitational and Kinetic Energy Calculations

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Date:	Sunday, 7 September 2025, 6:43 PM

Lesson 3 Gravitational and Kinetic Energy Calculations

Table of contents

Introduction

The kinetic energy of an object depends on more factors than just the gravitational potential energy of an object.

The kinetic energy of an object depends on more factors than just the gravitational potential energy of an object.

Targets

Watch This

Significant Digits

https://adlc.wistia.com/medias/p8ku1god3z

Gravitational Potential Energy

Gravitational potential energy can be found in many natural and technological systems.

Gravitational potential energy can be found in many natural and technological systems.

Examples

Read This

Practice Questions

Virtual Lab

Potential Energy on Shelves © 2017 Explore Learning

Potential Energy on Shelves © 2017 Explore Learning

Kinetic Energy

Kinetic energy can be found in many natural and technological systems.

Kinetic energy can be found in many natural and technological systems.

Examples

Read This

Practice Questions

Watch This

A Great Scientist Risks His Life Explaining Potential and Kinetic Energy © YouTube Energy Unearthed

Importance of Gravitational and Kinetic Calculations

Calculating the gravitational potential energy and kinetic energy of an object requires an understanding of many factors.

Calculating the gravitational potential energy and kinetic energy of an object requires an understanding of many factors.

Watch This

The Story of Kinetic and Potential Energy © YouTube cassiw2

The Story of Kinetic and Potential Energy © YouTube cassiw2

3.2 Assignment

Unit 3 Assignment Lessons 2-4

Lesson 3 Gravitational and Kinetic Energy Calculations

Table of contents

Introduction

The kinetic energy of an object depends on more factors than just the gravitational potential energy of an object.

The kinetic energy of an object depends on more factors than just the gravitational potential energy of an object.

Targets

Watch This

Significant Digits

https://adlc.wistia.com/medias/p8ku1god3z

Gravitational Potential Energy

Gravitational potential energy can be found in many natural and technological systems.

Gravitational potential energy can be found in many natural and technological systems.

Examples

Read This

Practice Questions

Virtual Lab

Potential Energy on Shelves © 2017 Explore Learning

Potential Energy on Shelves © 2017 Explore Learning

Kinetic Energy

Kinetic energy can be found in many natural and technological systems.

Kinetic energy can be found in many natural and technological systems.

Examples

Read This

Practice Questions

Watch This

A Great Scientist Risks His Life Explaining Potential and Kinetic Energy © YouTube Energy Unearthed

PWNs7i4rEWA

Importance of Gravitational and Kinetic Calculations

Calculating the gravitational potential energy and kinetic energy of an object requires an understanding of many factors.

Calculating the gravitational potential energy and kinetic energy of an object requires an understanding of many factors.

Watch This

The Story of Kinetic and Potential Energy © YouTube cassiw2 7K4V0NvUxRg

The Story of Kinetic and Potential Energy © YouTube cassiw2

3.2 Assignment

Unit 3 Assignment Lessons 2-4

The Story of Kinetic and Potential Energy © YouTube cassiw2