SPH4C/SPH3U
Findlay

Energy

Energy can be defined as the capacity to work or to accomplish a task.

Example: burning fuel allows an engine to do the work of moving a car.

Forms of Energy

Radiant Energy

Components of the electromagnetic spectrum have characteristics of waves, such as wavelengths, frequencies, and energies; they travel in a vacuum at the speed of light (3.0 x 10 8 m/s).

Forms of Energy

Kinetic Energy

Energy of motion, every moving object has this energy

Forms of Energy

Gravitational Potential Energy

A raised object has stored energy due to its position above some reference level

Forms of Energy

Elastic Potential Energy

Is stored in objects that are stretched or compressed.

Forms of Energy

Chemical Potential Energy

In chemical reactions, new molecules are formed and chemical potential energy is released or absorbed.

Forms of Energy

Nuclear Potential Energy

The nucleus of every atom contains energy.

Nuclear fission – The breaking down of atoms
 Nuclear fusion – the joining of atoms

Forms of Energy

Electrical Potential Energy

Electrons in a electric circuit can transfer energy to the components of the circuit.

Forms of Energy

Thermal Energy

The more rapidly atoms and molecules move, the greater their total thermal energy.

Forms of Energy

Sound Energy

Produced by vibrations; the energy travels by waves through a material to the receiver.

Energy Transformations

The 9 forms of energy listed above are able to change from one to another; this change is called energy transformation . A energy transformation equation can be used to summarize the changes in a transformation. Example: a microwave

Electrical energy  Radiant Energy  Thermal
Energy


A device used to transform energy for a specific purpose.

Mechanical Work

Mechanical work is done on an object when a force displaces the object in the direction of the force or a component of the force.

Work is not energy itself, but rather it is a transfer of mechanical energy.

Mechanical Work

The mechanical work, W, done by a force on an object is the product of the force, F, and the displacement, Δ𝑑 ,
𝑊 = 𝐹Δ𝑑

The magnitude of the force must be constant.

The force and displacement must be in the same direction.

Is this an example of work?

A teacher applies a force to a wall and becomes exhausted.

A book falls of a table and free falls to the ground.

A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed.

Example: Pushing a cart
How much mechanical work does a store manager do on a grocery cart if she applies a force with magnitude 25 N in the forward direction and displaces the cart 3.5 m in the same direction?
𝐹 𝑎𝑝𝑝
= 25 N forward
Δ m forward
𝑊 = ?
𝑊 = 𝐹Δ𝑑
𝑊 = 25 N 3.5
m
𝑊 = 88 Nm = 88 J
∴ the work done to move the cart is 88 J .

A curler applies a force of 15 N on a curling stone and accelerates the stone from rest to a speed of 8.00 m/s in 3.5 s.
Assuming that the ice surface is level and frictionless, how much mechanical work does the curler do on the stone?
𝐹 𝑎𝑝𝑝 𝑣 𝑣 𝑓 𝑖
= 15
= 0 m/s
= 8.00
N m/s
Δ𝑡 = 3.5
s
𝑊 = ?
𝑣 𝑖 𝑣 𝑓
Δ𝑡
2
0 m/s + 8.00
m/s
2
Δ m
3.5
s
𝑊 = 𝐹Δ𝑑
𝑊 = 15 N 14 m
𝑊 = 210 J
∴ the work done by the curler is 210 J .

Negative Work

If the force is opposite in direction to the displacement, the work done is negative.
𝑊 = −𝐹Δ𝑑

Consider the previous example of the shopping cart but this time a second employee exerts a horizontal force in the opposite direction to the 25 N force.

Since the force is in the opposite direction to the displacement, the work done by the second employee on the carts is negative.

Work Done by Friction

Negative work can also occurs with kinetic friction because the force of kinetic friction always happens in the direction opposite to the direction of motion of the object.
𝑊 = −𝐹 𝑘
Δ𝑑

𝐹 𝑘 is the magnitude of the force of Kinetic Friction.

Example
A toboggan carrying two children (total mass 85kg) reaches its maximum speed at the bottom of a hill. It then glides to a stop 21 m along a horizontal surface.
The coefficient of friction between the toboggan and the snowy surface is 0.11.
A.
Calculate the magnitude of the force of kinetic friction acting on the toboggan
B.
Calculate the work done by the force of kinetic friction on the toboggan

Example
A.
𝐹 𝑘
𝐹 𝑘
𝐹 𝑘 𝑚 = 85 kg 𝑔 = 9.8 N/kg 𝜇 𝑘
𝐹 𝑘
= 0.11
= ?
= 𝜇 𝑘
𝐹
𝑁
= 𝜇 𝑘 𝑚𝑔
= 0.11 85 9.8
= 92 N
B.
Δ𝑑 = 21 m
𝑊 = ?
𝑊 = −𝐹 𝑘
Δ𝑑
𝑊 = − 92 21
𝑊 = −1.9 × 10 3 J

Friction

The work done by friction has been transformed into thermal energy . This is observed as a increase in temperature.

Zero Work


Sometimes an object can experience a force, a displacement, or both, yet no work is done on the object.
Example: Holding a box in your hands. The box is not moving so no work is done so the displacement is zero.

A puck on an air hockey table is moving but it does not have force acting parallel to the movement as friction is negligible.

Positive, Negative, or Zero?


An object may experience a force in one direction while it moves in a different direction. This occurs when a person pulls on a suitcase with wheels and a handle.
𝐹 𝑎𝑝𝑝
𝐹 𝑦 𝜃
𝐹 𝑥




The applied force makes an angle, Θ , with the horizontal displacement. The force acting in the same direction as the displacement is the horizontal component of the applied force.
𝑊 = 𝐹Δ𝑑
𝑊 = 𝐹 𝑥
Δ𝑑
𝑊 = 𝐹 𝑎𝑝𝑝 cos 𝜃 Δ𝑑
This force is the only force that causes the suitcase to move along the floor. The work done by this force is,
𝑊 = 𝐹𝑐𝑜𝑠 Θ Δ𝑑
The vertical force, 𝐹 𝑦
, is perpendicular to the displacement and does not do work on the suitcase.

Example

Calculate the mechanical work done by a custodian on a vacuum cleaner if the custodian exerts an applied force of
50.0 N on the vacuum hose and the hose makes a 30.0
o with the floor. The vacuum cleaner moves 3.00 m to the right on a level, flat surface.


In many cases, objects can experience many forces at a time. The total work done, 𝑊 𝑛𝑒𝑡
, on the object is the sum of the work done by all of the forces acting on the object.
𝑊 𝑛𝑒𝑡
= 𝑊
1
+ 𝑊
2
+ ⋯ = 𝐹 𝑛𝑒𝑡 cos(Θ) Δ𝑑

Example
A shopper pushes a shopping cart on a horizontal surface with horizontal applied force of 41.0 N for
11.0 m. The cart experiences a force of friction of
35.0 N. Calculate the total mechanical work done on the shopping cart.

SPH4C/SPH3U
Findlay

Work





Energy is the capacity to do work and provides objects with the ability to do work.
Work is not energy itself, but rather a transfer of energy.
A force does work on an object if it causes the object to move.
Work is always done on an object and results in a change in the object.
The work done is equal to the change in energy.

Mechanical Energy

The mechanical work on an object is the amount of mechanical energy transferred to that object by a force.
𝑊 = Δ𝐸

The mechanical energy of an object is that part of its total energy which is subject to change by mechanical work.

Kinetic and Potential Energy

Kinetic energy is the energy of motion and
potential energy is the energy to, potentially, do something else.

Kinetic Energy

Energy due to the motion of an object.

Kinetic Energy

1.
2.
Objects energy depends on two factors:
Mass
Speed

When a force is applied to accelerate an object from speed 𝑣
1 written as, to speed 𝑣
2
, the work done on the object can be
𝑊 = 𝐹Δ𝑑 = 𝑚𝑎Δ𝑑

Change in Energy


We are interested in the kinetic energy, which means we want to see how the speed affects the energy.
Using the following equations from kinematics, we can find the change in kinetic energy.
𝑎 =
Δ𝑑 = 𝑣 𝑎𝑣
Δ𝑣
Δ𝑡
= 𝑣
2
Δ𝑡 =
− 𝑣
2
1 𝑣
1
Δ𝑡
+ 𝑣
2
Δt

Change in Energy
𝑊 = 𝑚𝑎Δ𝑑 = 𝑚 𝑣
2
− 𝑣
1
Δ𝑡 𝑣
1
+ 𝑣
2
2
Δ𝑡
𝑊 =
1
2 𝑚𝑣 2
2
−
1
2 𝑚𝑣 2
1
= Δ𝐸 𝑘
A change in speed represents a change in kinetic energy and the work done to change the speed represents a transfer of kinetic energy.

Kinetic Energy
The kinetic energy can be found from,
𝐸 𝑘
=
1
2 𝑚𝑣 2
Where



𝐸 𝑘 is the kinetic energy in joules ( J ) 𝑚 is the mass in kg 𝑣 is the speed in m/s

Example

Example
An object of mass 5.0 kg is travelling at a speed of 4.0 m/s.
27 J of work is done to increase the speed of the object.
Calculate its final kinetic energy and final speed.
𝑚 = 5.0 kg 𝑣
1
= 4.0 m/s
𝑊 = 27 J
𝐸 𝑘2 𝑣
2
= ?
= ?
𝑊 = 𝐸 𝑘2
− 𝐸
𝐸
𝐸 𝑘2 𝑘2 𝑘1 or 𝐸 𝑘2
= 27 J +
= 67 J
= 𝑊 + 𝐸
1 𝑘1
5.0 4.0
2
2 𝑣
2
=
2𝐸 𝑘2 𝑚 𝑣
2
=
2 67
5.0
𝑣
2
= 5.2 m/s

Gravitational Potential Energy

The type of energy that an object possess because of its position above some level.

The energy is called potential because it can be stored and used at a lower level for work .

When an object falls, its potential energy is transformed into kinetic energy as its speed increases.

Gravitational Potential Energy


Since the force ( 𝐹 ) required to lift an object without accelerating is the same as the objects weight ( 𝑚𝑔 ), the energy ( 𝐸 𝑔
) required to lift an object is the same as its potential energy from the height it was lifted from.
Since 𝐹 𝑔
= 𝑚𝑔 , we can make a common equation for gravitational potential energy.
𝐸 𝑔
= 𝐹 𝑔
Δh = mgΔh

Energy is measured in joules and height is measured in meters.

Gravitational Potential Energy

The potential energy in a system must be based against a Reference Level – The level to which the object may fall.

It is important to note that when answering questions about relative potential energy, it is important to state the reference level.

Example
In the sport of pole vaulting, the jumper’s point of mass centralization, called the centre of mass, must clear the pole. Assume that a 59 kg jumper must raise the centre of mass 1.1 m off the ground to 4.6 m off the ground. What is the jumper’s gravitational potential energy at the top of the bar relative to the point at which the jumper started the jump?

Example ℎ = 4.6 − 1.1 = 3.5 m 𝑚 = 59 kg 𝑔 = 9.8 N/kg
𝐸 𝑔
= ?
𝐸 𝑔
𝐸 𝑔
𝐸 𝑔
= 𝑚𝑔Δℎ
= 59 9.8 3.5
= 2.0 × 10 3 J

Mechanical Energy

The sum of gravitational potential energy and kinetic energy is called mechanical energy .
𝐸 𝑚𝑒𝑐ℎ
= 𝐸 𝑘
+ 𝐸 𝑔