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Lesson 4 Gravitational and Potential Energy

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Course: Science 10 [5 cr] - AB Ed copy 1
Book: Lesson 4 Gravitational and Potential Energy
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Date: Sunday, 7 September 2025, 6:44 PM

Table of contents

  Introduction

Mechanical energy is the energy that an object has due to its motion or position.



C4.1 Volleyball
Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position). Objects have mechanical energy if they are in motion or if they are held at a position above the surface of Earth.

Some examples of mechanical energy are
  • a child held on his parents shoulders due to his or her position above the ground
  • a moving car due to its speed and motion
  • a thrown football due to its speed and position above the ground
  • a hand weight lifted high above a person’s head due to its position above the ground

  Target

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

  • relate how kinetic energy always changes into another type of energy due to energy conservation

  Watch This

Conservation of Energy © YouTube Bozeman Science


This video reviews concepts covered so far in this section of the unit. Energy can neither be created nor destroyed but may be transferred. Energy comes in many forms (including chemical, mechanical, light, electrical, and thermal). This lesson looks further at mechanical energy. This will help get you in the right mindset for this lesson.


  What is Mechanical Energy?

An object that has mechanical energy is able to do work.


Mechanical energy is often defined as the ability to do work. Remember, this is also the definition of energy. Any object that has mechanical energy, whether it is in the form of gravitational potential energy or kinetic energy, is able to do work. Mechanical energy allows an object to apply a force to another object in order to cause it to be moved.

The total amount of mechanical energy is the sum of the gravitational potential energy and the kinetic energy.


«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi mathcolor=¨#FFFFFF¨»mechanical«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»energy«/mi»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mi mathcolor=¨#FFFFFF¨»kinetic«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»energy«/mi»«mo mathcolor=¨#FFFFFF¨»+«/mo»«mi mathcolor=¨#FFFFFF¨»potential«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»energy«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub mathcolor=¨#FFFFFF¨»«mi mathcolor=¨#FFFFFF¨»E«/mi»«mi»m«/mi»«/msub»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«msub mathcolor=¨#FFFFFF¨»«mi mathcolor=¨#FFFFFF¨»E«/mi»«mi»k«/mi»«/msub»«mo mathcolor=¨#FFFFFF¨»+«/mo»«msub mathcolor=¨#FFFFFF¨»«mi mathcolor=¨#FFFFFF¨»E«/mi»«mi»p«/mi»«/msub»«/mtd»«/mtr»«mtr»«mtd»«msub mathcolor=¨#FFFFFF¨»«mi mathcolor=¨#FFFFFF¨»E«/mi»«mi»m«/mi»«/msub»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mfrac mathcolor=¨#FFFFFF¨»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi mathcolor=¨#FFFFFF¨»m«/mi»«msup mathcolor=¨#FFFFFF¨»«mi mathcolor=¨#FFFFFF¨»v«/mi»«mn»2«/mn»«/msup»«mo mathcolor=¨#FFFFFF¨»+«/mo»«mi mathcolor=¨#FFFFFF¨»m«/mi»«mi mathcolor=¨#FFFFFF¨»g«/mi»«mi mathcolor=¨#FFFFFF¨»h«/mi»«/mtd»«/mtr»«/mtable»«/math»

C4.2 Bowling ball rolling

Also, remember that work is a measurement of energy in a system. So, the work done can also be transformed into gravitational potential energy or kinetic energy.

The law of conservation of energy states that the total amount of energy in any given situation remains constant. In a system, an object’s gravitational potential energy can be converted to kinetic energy and vice versa. The assumption is made that energy will be transferred from one form to another with no energy lost. That is why in any system, the amount of gravitational potential energy is equal to the amount of kinetic energy.

For an object falling a distance"h" from rest:


«math» «mi mathcolor=¨#FFFFFF¨»m«/mi» «mi mathcolor=¨#FFFFFF¨»g«/mi» «mi mathcolor=¨#FFFFFF¨»h«/mi» «mo mathcolor=¨#FFFFFF¨»=«/mo» «mi mathcolor=¨#FFFFFF¨»F«/mi» «mi mathcolor=¨#FFFFFF¨»d«/mi» «mo mathcolor=¨#FFFFFF¨»=«/mo» «mfrac mathcolor=¨#FFFFFF¨» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi mathcolor=¨#FFFFFF¨»m«/mi» «msup mathcolor=¨#FFFFFF¨» «mi mathcolor=¨#FFFFFF¨»v«/mi» «mn»2«/mn» «/msup» «/math»


Examples

  1. A 150 g ball has 12.2 J of gravitational potential energy as it is held 6.8 m above the ground. Determine the amount of kinetic energy the ball has just as it is about to hit the ground after being dropped.


As the ball falls, the gravitational potential energy gets converted into kinetic energy. If the gravitational potential energy was 12.2 J, the kinetic energy is 12.2 J.
  1. A 14.3 kg chunk of ice falls from a roof that is 8.5 m above the ground. Calculate the ice chunk’s kinetic energy as it reaches the ground.

    As the ice chunk falls, the gravitational potential energy gets converted into kinetic energy.

    Determine the Ep of the ice chunk. 


Step 1:  List the variables

m = 14.3 kg

h = 8.5 m

g = 9.81 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«/math»  (this one is easy to forget because it is not listed in the questions.  Find it in the Data Booklet.

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mo»?«/mo»«/math»

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

              «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/math»

              «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/math»

Step 3:  Substitute into the formula

There are two ways to consider this problem.

Method 1:  The strength of this method is simplicity.  It works in this situation where energy is 100% converted.

Recognise that the potential energy of the ice on the roof is 100 % converted to kinetic energy just before the ice hits the ground.

     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«mn»14«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»8«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/math»

Method 2:  Mechanical energy is conserved, so consider the mechanical energy at two time. This is not really different.  The advantage is it can be applied in more complicated situations.

Mechanical energy is conserved, so think about mechanical energy before and after.

Before
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mn»14«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»8«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»1192«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi»J«/mi»«/math»

 After

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mspace linebreak=¨newline¨»«/mspace»«mn»1192«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi»J«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/math»

Step 4:  Calculate the answer

     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»1192«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi»J«/mi»«/math»  .  The smallest number of significant digits in the question is 2 (8.5 m), so your answer must be expressed with two significant digits.  To do this you need to use scientific notation

     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»1«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi»J«/mi»«/math»



  1. A girl with a mass of 55 kg stands at the top of a 10.0 m platform diving tower at the local swimming pool. She steps off and falls to the water below. Calculate the girl’s speed as she contacts the water.

    As the girl falls, the gravitational potential energy gets converted into kinetic energy. Ep = Ek due to the conservation of energy.

 Step 1: List the variables

      m = 55 kg

      h = 10.0 m

     g = 9.81 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«/math»

Step 2:  Identify the correct formula

This questions involved and change in mechanical energy, so the the appropriate formula is:

     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/math»

Step 3:  Substitute values into the formula.

     There are two ways to think about this problem.  the first method is simple and works when energy is 100% converted.  the second method works in every situation.

Method 1:  Recognizing the potential energy is 100% converted to Kinetic energy

null

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160;§#160; 55«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»10«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mspace linebreak=¨newline¨»«/mspace»«/math»

Method 2: Mechanical energy is conserved.  consider the mechanical energy at two times

        Before

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»10«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»5395«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»J«/mi»«/math»
       After
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mn»0«/mn»«mspace linebreak=¨newline¨»«/mspace»«mn»5395«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»J«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»x«/mi»«mo»§#160;«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

Step 4:  Calculate the answer.

     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»5395«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»J«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mspace linebreak=¨newline¨»«/mspace»«mfrac»«mrow»«mn»5395«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»J«/mi»«/mrow»«mrow»«mn»27«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«/mrow»«/mfrac»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mspace linebreak=¨newline¨»«/mspace»«mn»196«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«msup»«mi»m«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mo»/«/mo»«mo»§#160;«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mspace linebreak=¨newline¨»«/mspace»«msqrt»«mn»196«/mn»«mo».«/mo»«mn»2«/mn»«msup»«mi»m«/mi»«mn»2«/mn»«/msup»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«/msqrt»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mo»§#160;«/mo»«msqrt»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/msqrt»«mspace linebreak=¨newline¨»«/mspace»«mo»§#160;«/mo»«mo»§#160;«/mo»«mn»14«/mn»«mo».«/mo»«mn»0071«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«mi»s«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mi»v«/mi»«/math»

Expressed with two significant digits because 55 kg has only two, the answer becomes 14 m/s.

     v = 14 m/s

4.  A 2.5 m long canon is aimed straight up.  It applies an average force of 1300 N on a cannonball.  What will be the maximum height of the cannonball?  (Assume no energy is lost)

 Step 1:  List the variables.

d = 2.5 m     F = 1300 N      m = 3.5 kg           h = ?    

Step 2:  Choose the correct formula.

      In this situation work is done on the cannonball to give it kinetic energy, which then is changed to potential energy.

W = Fd                «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/math»

Step 3:  Substitute and calculate.  This is a two step problem.

 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»W«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mi»f«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mi»d«/mi»«mspace linebreak=¨newline¨»«/mspace»«mi»W«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»1300«/mn»«mo»§#160;«/mo»«mi»N«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mspace linebreak=¨newline¨»«/mspace»«mi»W«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»3261«/mn»«mo».«/mo»«mn»825«/mn»«mo»§#160;«/mo»«mi»J«/mi»«mspace linebreak=¨newline¨»«/mspace»«/math»

 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»W«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«mn»3261«/mn»«mo».«/mo»«mn»825«/mn»«mo»§#160;«/mo»«mi»J«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»3«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mi»x«/mi»«mo»§#160;«/mo»«mi»h«/mi»«mspace linebreak=¨newline¨»«/mspace»«mfrac»«mrow»«mn»3261«/mn»«mo».«/mo»«mn»825«/mn»«mo»§#160;«/mo»«mi»J«/mi»«/mrow»«mrow»«mn»34«/mn»«mo».«/mo»«mn»335«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«msup»«mi»s«/mi»«mn»2«/mn»«/msup»«/mrow»«/mfrac»«mo»§#160;«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mi»h«/mi»«mspace linebreak=¨newline¨»«/mspace»«mn»95«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mi»h«/mi»«/math»

 


Step 4:  Express your answer.

The least number of significant digits in the question is two, so the asnower should be expressed with two significant digits

h = 95 m

  Read This

Please read pages 183 to 185 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on mechanical energy and the law of conservation of energy and the calculations used in these two concepts. Remember, if you have any questions or do not understand something, ask your teacher!

  Practice Questions

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.

  1. A 31.2 kg child is jumping on a trampoline. If the child jumps vertically at an initial speed of 2.47 m/s, calculate how high the child will rise.


Due to the conservation of energy, the gravitational potential energy of the child is equal to the kinetic energy of the child.

    Step 1: List the variables.

    «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»31«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/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»47«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«/mtable»«/math»
    Step 2: Identify the correct formula and rearrange if necessary.

    «math»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»=«/mo»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mspace linebreak=¨newline¨»«/mspace»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

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

    «math» «mfrac» «mrow» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «/mrow» «mrow» «mi»m«/mi» «mi»g«/mi» «/mrow» «/mfrac» «mo»=«/mo» «mfrac» «mstyle displaystyle=¨true¨» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/mstyle» «mrow» «mi»m«/mi» «mi»g«/mi» «/mrow» «/mfrac» «/math»

    Now, cancel the like terms.

    «math» «mi»h«/mi» «mo»=«/mo» «mfrac» «mrow» «mstyle displaystyle=¨true¨» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/mstyle» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/mrow» «mi»g«/mi» «/mfrac» «/math»
    Step 3: Substitute the values into the formula.

    «math»«mi»h«/mi»«mo»=«/mo»«mfrac»«mrow»«mfenced»«mstyle displaystyle=¨true¨»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mstyle»«/mfenced»«msup»«mfenced»«mrow»«mn»2«/mn»«mo».«/mo»«mn»47«/mn»«mstyle displaystyle=¨true¨»«mo»§#160;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mstyle»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mrow»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mstyle displaystyle=¨true¨»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mstyle»«/mrow»«/mfenced»«/mfrac»«/math»
    Step 4: Calculate the answer.

    h = 0.310 9...m

    The answer must be rounded to thee significant digits.

    The child will rise 0.311 m.
  1. A person is playing darts and throws a 22.3 g dart with a horizontal speed of 6.13 m/s at the dartboard. The total mechanical energy of the dart is 0.765 J. Calculate the gravitational potential energy of the dart.

This question is looking for gravitational potential energy, so you can use the relationship that mechanical energy is equal to the gravitational potential energy added to the kinetic energy.

    Step 1: List the variables.

    «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»m«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»756«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»22«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»22«/mn»«mo».«/mo»«mn»3«/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»0«/mn»«mo».«/mo»«mn»022«/mn»«mo»§#160;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«/mtable»«/math»

    Since m is measured in kilograms, we have to convert our units of g to kg.

    «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mo».«/mo»«mn»13«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«/mtable»«/math»
    Step 2: Identify the correct formula and rearrange if necessary.

    Em = Ep + Ek

    To isolate Ep by itself, subtract both sides by Ek. To move Ek to the other side, you need to use the opposite operation.

    «math» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «msub» «mi»E«/mi» «mi»m«/mi» «/msub» «mo»-«/mo» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mfenced» «mrow» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»+«/mo» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mrow» «/mfenced» «mo»-«/mo» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mtd» «/mtr» «mtr» «mtd» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «msub» «mi»E«/mi» «mi»m«/mi» «/msub» «mo»-«/mo» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mtd» «/mtr» «/mtable» «/math»
    Remember that «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».

    So, «math» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «msub» «mi»E«/mi» «mi»m«/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»p«/mi»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»765«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«mo»-«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»022«/mn»«mo»§#160;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»6«/mn»«mo».«/mo»«mn»13«/mn»«mo»§#160;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/math»
    Step 4: Calculate the answer.

    Ep = 0.795 J – 0.418 92...J
    Ep = 0.346 0...J

    The answer must be rounded to three significant digits.

    The potential energy of the dart is 0.346 J.

  Virtual Lab

Energy Skate Park © 2017 PhET
Work through this simulation activity to reinforce what you learned about mechanical energy, gravitational potential energy, and kinetic energy in a system.

You will manipulate a skate boarder at a skate park to investigate energy conversions that occur in a mechanical system.



  1. Click on the play button to open the activity. This activity can also be accessed at https://quick.adlc.ca/skatepark Click on “Intro.”
  2. Click on “Bar Graph.” Place the skater at the top of the left side of the track. Watch as the skater moves across the track and how it changes the levels of potential and kinetic energy. (You can slow or pause the skater down at the bottom of the simulator window to help you observe).
  1. When is the gravitational potential energy the highest?

    When the skater is at the top of either side of the track.
    The lowest?

    When the skater is at the bottom of the track.
  1. When is the kinetic energy the highest?

    When the skater is at the bottom of the track.
    The lowest?

    When the skater is at the top of either side of the track.
  1. When are the kinetic and gravitational potential energy levels the same?

    When the skater is halfway through the path on either side of the track.
  2. What do you observe about the total energy value? (This is the mechanical energy of the system).

    The total (mechanical) energy does not change through the movement of the skater.
  3. Place the skater halfway down on the left side of the track. Watch as the skater moves across the track and how it changes the levels of potential and kinetic energy at this new starting height.

  4. How did the total energy value change when you moved the skater to the new starting point?

    The total (mechanical) energy has decreased with the movement of the skater.
  5. Click on “Speed.” Place the skater at the top of the left side of the track. Watch as the skater moves across the track and how it changes the levels of potential and kinetic energy and also the speed. (You can slow or pause the skater down at the bottom of the simulator window to help you observe).

  1. When is the speed the highest?

    When the skater is at the bottom of the track.
    The lowest?

    When the skater is at the top of either side of the track.
  2. Move “Mass” to “Small.”
  3. What do you observe about the total energy value? (This is the mechanical energy of the system).

    The total (mechanical) energy still does not change through the movement of the skater, but it has an overall smaller value.
  4. Move “Mass” to “Large.”
  5. What do you observe about the total energy value? (This is the mechanical energy of the system).

    The total (mechanical) energy still does not change through the movement of the skater, but it has an overall larger value.
  6. Please return to the top of this page and click on analysis to complete the analysis questions.
  1. Can you think of a scenario when the kinetic and gravitational potential energy of the skater could both be zero? Explain.

    Kinetic energy is energy of motion, so the only time a skater’s kinetic energy is zero is when it is not moving.

    Gravitational potential energy is energy of position relative to Earth’s surface, so Ep can be zero if it is on the surface of Earth.

    If the skater stops moving, or falls, and lies at the bottom of the track, then the skater’s kinetic and gravitational potential energy could both be zero.

  2. Use the law of conservation of energy to explain how you can predict the height that the skater will reach on the right side depending on where you release them on the left side.

    The skater will reach the same height on the right side as the height she was released on the left side. This is because as the skater’s kinetic and gravitational potential energies are being converted back and forth, energy is not created nor destroyed.

  Mechanical Energy

The law of conservation of energy drives situations involving mechanical energy.


C4.3 Pendulum
You have now looked at how the law of conservation of energy explains why mechanical energy in a system remains constant. Kinetic energy can be converted to gravitational potential energy and vice versa, and no energy is lost.

In the next section of this unit, we will look at how we measure energy in mechanical systems, both theoretically and practically.

  Watch This

Pendulum Waves © YouTube Harvard Natural Sciences Lecture Demonstrations 


This video clip shows an amazing visual of how 15 pendulums of increasing lengths appear to dance as energy conversions simultaneously occur.

The time of one complete cycle of the dance is 60 seconds. The length of the longest pendulum has been adjusted so that it completes 51 cycles in this 60-second period. The length of each successive shorter pendulum is adjusted so that it completes one more cycle in this time. So, the 15th pendulum (shortest) undergoes 65 cycles. When all 15 pendulums are started together, they quickly fall out of sync. However, after 60 seconds, they will all be back in sync again at that instant, ready to repeat the dance.


3.2 Assignment

Unit 3 Assignment Lessons 2-4

It is now time to complete the Lesson 3 portion of 3.2 Assignment. This assignment has two parts.

  1. Part 1 Written Portion: Select the preferred document type from the options below. Download and save the assignment on your desktop (or documents folder).

    PDF Document       
  2. Open and print this saved document.
  3. Record your responses in the appropriate textboxes.
  4. When you have completed the assignment, scan it and save it on your desktop (or documents folder).
  5. Once you have completed the written portion of your assignment, click on the button below to go to the submission page.

    Written Portion Submission Page
  6. Part 2 Online Portion: It is now time to complete the online portion of this assignment. Click on the button below to go to the online questions of this assignment.

    Online Questions

This assignment is worth ___% of your final grade.