Motion 2
Momentum and Energy
UNIT TWO PHYSICS
Momentum
Momentum can be defined as ‘mass in motion’.
If an object has a mass and is moving, then it has momentum.
Knowing about Momentum helps us to answer questions like:
◦ How difficult is it to stop a moving object?
◦ How difficult is it to make a stationary object move?
The answer to these questions depends on two things:
• The mass of the object
• How fast the object is going / How fast you want the object to go
Momentum
The product of these two characteristics is called momentum and can be
defined as:
𝑝 = 𝑚𝑣
Momentum is a
vector quantity
𝑚 is the mass of the object measured in 𝑘𝑔
𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟 𝑚/𝑠
𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1 or 𝑘𝑔 𝑚/𝑠
Momentum
𝑚 is the mass of the object measured in 𝑘𝑔
𝑝 = 𝑚𝑣
𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟
𝑚
𝑠
𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1
or 𝑘𝑔 𝑚/𝑠
eg1. Find the momentum of a car of mass 1500 𝑘𝑔, moving north at a velocity of 20 𝑚 𝑠 −1 ?
𝑝 = 𝑚𝑣
= 1500 𝑘𝑔 × 20 𝑚 𝑠 −1
= 3000 𝑘𝑔 𝑚 𝑠 −1 North
Momentum
𝑚 is the mass of the object measured in 𝑘𝑔
𝑝 = 𝑚𝑣
𝑣 is the velocity of the object, measured in 𝑚 𝑠 −1 𝑜𝑟
𝑚
𝑠
𝑝 is the momentum of the object, measured in 𝑘𝑔 𝑚 𝑠 −1
or 𝑘𝑔 𝑚/𝑠
eg2. A train with a momentum of 20 × 106 𝑘𝑔 𝑚/𝑠 travels west at a velocity of 120 km/hr.
What is the mass of the train?
𝑝 = 𝑚𝑣
𝑚=
=
𝑝
𝑣
20×106 𝑘𝑔 𝑚 𝑠 −1
(120 ÷ 3.6) 𝑚 𝑠 −1
= 600,000 𝑘𝑔 = 6 × 105 𝑘𝑔 west
Momentum & Impulse
•
Making an object stop or go requires a non-zero net force.
•
Recall Newtons second law,
•
𝐹𝑛𝑒𝑡 = 𝑚𝑎
Acceleration is change in velocity over a time interval
•
Combining these together gives: 𝐹𝑛𝑒𝑡 = 𝑚(
•
Rearranging gives: 𝐹𝑛𝑒𝑡 ∆𝑡 = 𝑚 ∆𝑣
∆𝑣
∆𝑡
∆𝑣
∆𝑡
)
This gives the impulse, 𝐹𝑛𝑒𝑡 ∆𝑡 and can be defined as the product of the net
force over the time interval which it acts.
Momentum & Impulse
Impulse = 𝐹𝑛𝑒𝑡 ∆𝑡 = 𝑚 ∆𝑣
= 𝑚(𝑣 − 𝑢)
= 𝑚𝑣 − 𝑚𝑢
= 𝑝𝑓 − 𝑝𝑖
Where 𝑝𝑓 = 𝐹𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑝𝑖 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
So we can also say that impulse
is the change in momentum, ∆𝑝
And we can say Net Force is the rate of change of momentum given by:
𝐹𝑛𝑒𝑡 =
∆𝑝
∆𝑡
∆𝑝 = Change in momentum
∆𝑡 = Change in time
eg1. A 200g netball hits a wall horizontally at a speed of 12 m/s and bounces back
in the opposite direction at speed of 10 m/s. The ball comes into contact with the
wall for 10 ms.
a) What is the change in momentum of the netball ball?
Assign initial direction on ball as positive
u = 12 m/s
v = 10 m/s
∆𝑝 = 𝑚∆𝑣
= 𝑚 𝑣−𝑢
= 0.200 × −10 − 12
= 0.200 × −22
= −4.4 𝑘𝑔 𝑚 𝑠 −1
b) What is the impulse on the netball?
Impulse on the tennis ball = change in momentum of the tennis ball
= −4.4 𝑘𝑔 𝑚 𝑠 −1
c) What is the magnitude of the force exerted by the wall on the netball?
𝐹𝑛𝑒𝑡 =
∆𝑝
∆𝑡
=
4.4 𝑘𝑔 𝑚 𝑠 −1
0.2
= 22 𝑁
Now Try
Momentum Worksheet
then…..Text Book Problems
Chapter 6
Questions 46, 47, 48, 49, 50, 51
Basketball Scenario
a) What is the initial momentum of the ball?
Assign initial direction on ball as positive
Vinitial = +10 m/s
b) What is the final momentum of the ball?
Assign final direction on ball as negative
c) What is the Impulse of the ball?
Vfinal = -7 m/s
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚𝑣
= 0.020 𝑘𝑔 × 10 𝑚𝑠 −1
= 0.20 𝑘𝑔 𝑚 𝑠 −1
𝑝𝑓𝑖𝑛𝑎𝑙 = 𝑚𝑣
= 0.020 𝑘𝑔 × −7 𝑚𝑠 −1
= −0.14 𝑘𝑔 𝑚 𝑠 −1
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = ∆𝑝 = 𝑝𝑓𝑖𝑛𝑎𝑙 − 𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −0.14 − 0.20
d) What is the magnitude of the Net force exerted by the wall on
the0.34
ball?
=−
𝑘𝑔 𝑚 𝑠 −1
𝐹𝑛𝑒𝑡
∆𝑝
0.34 𝑘𝑔 𝑚 𝑠 −1
=
=
= 34 𝑁
∆𝑡
10 × 10−3 𝑠
eg. A 20g ping pong ball hits a wall horizontally at a speed of 10 m/s, before bouncing
off the wall at a speed of 7 m/s. The ball is in contact with the ball for 10 mS
a) What is the initial momentum of the ball?
Assign initial direction on ball as positive Vinitial = +10 m/s
b) What is the final momentum of the ball?
Assign final direction on ball as negative
c) What is the Impulse of the ball?
Vfinal = -7 m/s
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚𝑣
= 0.020 𝑘𝑔 × 10 𝑚𝑠 −1
= 0.20 𝑘𝑔 𝑚 𝑠 −1
𝑝𝑓𝑖𝑛𝑎𝑙 = 𝑚𝑣
= 0.020 𝑘𝑔 × −7 𝑚𝑠 −1
= −0.14 𝑘𝑔 𝑚 𝑠 −1
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = ∆𝑝 = 𝑝𝑓𝑖𝑛𝑎𝑙 − 𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= −0.14 − 0.20
= − 0.34 𝑘𝑔 𝑚 𝑠 −1
d) What is the magnitude of the Net force exerted by the wall on the ball?
𝐹𝑛𝑒𝑡
∆𝑝
0.34 𝑘𝑔 𝑚 𝑠 −1
=
=
= 34 𝑁
∆𝑡
10 × 10−3 𝑠
Impulse using graphs
The force calculated in the previous example is the Average Force on the ball – The
force acting on the ball is changing, reaching maximum value when the ball is at its
smallest distance from the wall.
𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝐹𝑎𝑣 ∆𝑡
We can use the real Force v Time graphs to find the impulse.
This is given by the area under the graph.
Impulse - using graphs
We can use the real Force v Time graphs to find the impulse.
This is given by the area under the graph.
eg. For the Force – Time graph given,
calculate the Impulse.
Area under
1
graph = 𝑏ℎ
2
1
= ×
2
40 × 100
= 2000 𝑁𝑠
Impulse - using graphs
eg. The Force – Time graph shows the force on a 50kg skater as she skates from rest.
a) Calculate the Impulse for the skater for the 2-second interval shown on the graph.
Area under graph = A + B + C
1
1
=
𝑏ℎ + 𝑙 × 𝑤 + ( 𝑏ℎ)
=
2
1
×
2
2
1
2
1.1 × 400 + 0.9 × 200 + ( × 0.9 × 200)
= 220 + 180 + 90
= 490 𝑁𝑠
b) Estimate her speed after 2 seconds.
∆𝑡 = 2.0 𝑠
∆𝑝 = 490 𝑘𝑔 𝑚𝑠 −1
𝑚 = 50 𝑘𝑔
∆𝑝 = 𝑚∆𝑣
∆𝑝 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖
490 = 50 × 𝑣𝑓 − 50 × 0
490
𝑣𝑓 =
= 9.8 𝑚𝑠 −1
50
Impulse - using graphs
eg. The Force – Time graph shows the force on a 50kg skater as she skates from rest.
c) Estimate the speed of the skater after 1.1 seconds
∆𝑡 = 1.1 𝑠
∆𝑝 = 220 𝑘𝑔 𝑚𝑠 −1
𝑚 = 50 𝑘𝑔
∆𝑝 = 𝑚∆𝑣
∆𝑝 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖
220 = 50 × 𝑣𝑓 − 50 × 0
220
𝑣𝑓 =
= 4.4 𝑚𝑠 −1
50
d) What is her acceleration over the first 1.1 seconds?
a=
=
∆𝑣
∆𝑡
4.4
1.1
= 4 𝑚𝑠 −2
Now Try
Worksheet - Impulse using graphs
then…..Text Book Problems
Chapter 6
Questions 46, 47, 48, 49, 50, 51, 52
Law of Conservation of Momentum
The law of conservation of momentum states that….
The total momentum of a closed system does not change.
This means that if two objects collide, the total momentum of the objects before
the collision is the same as the total momentum of the objects after the collision.
Total Momentum means the combined momentum of Object 1 + Object 2.
Law of Conservation of Momentum
After a collision, each object may not have the same momentum as it did initially,
but the combined momentum of the 2 objects will still be the same as before.
So if one object loses momentum in a collision, the other object gains momentum.
eg. A 2 kg cart is moving east at a constant velocity of 3 𝑚𝑠 −1 , when a 5kg brick is
placed on it (from rest) and continues to move east. What is the velocity of the cart and
brick system as it moves away (ignoring resistance forces) ?
Law of Conservation of Momentum
eg. A 3 kg cart is moving east at a constant velocity of 8 𝑚𝑠 −1 , when a 5kg brick is
placed on it (from rest) and continues to move east. What is the velocity of the cart and
brick system as it moves away (ignoring resistance forces) ?
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑓𝑖𝑛𝑎𝑙
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚𝑓𝑖𝑛𝑎𝑙 𝑣𝑓𝑖𝑛𝑎𝑙
2 𝑘𝑔 × 8 𝑚𝑠 −1 = 8 𝑘𝑔 × 𝑣𝑓
16 = 8 𝑣𝑓
𝑣𝑓 =
16
8
= 2 𝑚𝑠 −1
This type of collision is called
an inelastic collision – where
the objects move off
together after the collision
Modelling Collisions
Inelastic Collisions
Are collisions between two objects that result in the objects moving off
together, at a new combined mass and velocity.
Momentum is conserved.
This means that (as shown)……
Initial Momentum of the objects individually = Final Momentum of the objects together
𝑝𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑝𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑂𝐵𝐽𝐸𝐶𝑇 1 & 2 𝑓𝑖𝑛𝑎𝑙
Modelling Collisions
Elastic Collisions
Are collisions between two objects that result in the objects bouncing off each other.
Momentum is conserved.
This means that……
Initial total Momentum of the objects = Final total Momentum of the objects
𝑝𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑝𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑂𝐵𝐽𝐸𝐶𝑇 1 𝑓𝑖𝑛𝑎𝑙 + 𝑝𝑂𝐵𝐽𝐸𝐶𝑇 2 𝑓𝑖𝑛𝑎𝑙
Elastic Collisions
eg. A 1000kg car travelling at 20 𝑚𝑠 −1 hits a 3000kg stationary truck.
The truck moves forward at a 10 𝑚𝑠 −1 velocity and the car rebounds off the truck, in
the direction opposite to what it was initially travelling in.
What is the velocity of the car after the collision?
𝑝𝑐𝑎𝑟 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) + 𝑝𝑡𝑟𝑢𝑐𝑘 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 𝑝𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙) + 𝑝𝑡𝑟𝑢𝑐𝑘 (𝑓𝑖𝑛𝑎𝑙)
20 𝑚𝑠 −1 × 1000 𝑘𝑔
+
0
= 1000𝑣𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙) + 10 𝑚𝑠 −1 × 3000 𝑘𝑔
20000 = 1000𝑣 + 30000
−10000 = 1000𝑣
𝑣𝑐𝑎𝑟 (𝑓𝑖𝑛𝑎𝑙)
−10000
=
= −10 𝑚𝑠 −1
1000
For more examples click the link:
http://www.physicsclassroom.com/Class/momentum/u4l2d.cfm
Now Try
Cart Simulation Activity
Click the link to access the simulator:
http://www.physicsclassroom.com/Physics-Interactives/Momentum-andCollisions/Collision-Carts/Collision-Carts-Interactive
Have a play with the simulator then use the worksheet to investigate and
answer the problems
Now Try
Chapter 7
Questions 2a, 3, 4, 5
Conservation of Momentum
Eg1. A 10kg cart is travelling at 5m/s before a 7kg mass (initially at rest) is carefully
placed on top of it. Ignoring friction:
a) What is the initial momentum (combined) before the collision?
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚1 𝑣1 + 𝑚2 𝑣2 = 10 × 5 + 0 = 50 𝑘𝑔 𝑚/𝑠
b) What is the combined momentum after the collision?
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑓𝑖𝑛𝑎𝑙 = 50 𝑘𝑔 𝑚/𝑠
c) What is the velocity of the cart after the collision?
50
𝑝 = 𝑚𝑣 → 50 = 17𝑣 → 𝑣 =
= 2.94 𝑚/𝑠
17
Eg. A ball of mass 3kg moving at 5 m/s, collides elastically with a ball of mass 1kg, moving
at 3m/s in the same direction. The speed of the 3kg ball after the collision is 4m/s.
a) What is the total combined momentum of the two balls before the collision.
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚1 𝑣1 + 𝑚2 𝑣2 = 3 × 5 + 1 × 3 = 15 + 3 = 18 𝑘𝑔 𝑚/𝑠
b) What is the combined momentum after the collision? 𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑓𝑖𝑛𝑎𝑙 = 18 𝑘𝑔 𝑚/𝑠
c) What is the velocity of the 1kg ball after the collision?
𝑝𝑓𝑖𝑛𝑎𝑙 = 𝑚1 𝑣1 + 𝑚2 𝑣2
18 = 3 × 4 + 1𝑣
18 = 12 + 𝑣
𝑣 = 18 − 12 = 6 𝑚/𝑠
d) What is the change in momentum of the 3 kg ball? ∆𝑝 = 𝑚∆𝑣 = 3 × 5 − 4
= 3 × 1 = 3 𝑘𝑔 𝑚/𝑠
Conservation of Momentum
A ball of mass 3kg moving at 5 m/s, collides elastically with a ball of mass 1kg, moving
at 3 m/s in the same direction. The speed of the 3kg ball after the collision is 4 m/s.
a) What is the total combined momentum of the two balls before the collision.
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑚1 𝑣1 + 𝑚2 𝑣2 = 3 × 5 + 1 × 3 = 15 + 3 = 18 𝑘𝑔 𝑚/𝑠
b) What is the combined momentum after the collision?
𝑝𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑝𝑓𝑖𝑛𝑎𝑙 = 18 𝑘𝑔 𝑚/𝑠
c) What is the velocity of the 1kg ball after the collision?
𝑝𝑓𝑖𝑛𝑎𝑙 = 𝑚1 𝑣1 + 𝑚2 𝑣2 → 18 = 3 × 4 + 1𝑣 → 18 = 12 + 𝑣
𝑣 = 18 − 12 = 6 𝑚/𝑠
Now Do
Prac Activity Brick and Cart collision
Work
• When a force acts upon an object to cause a displacement of the object, it is
said that work was done upon the object.
• There are three key ingredients to work - force, displacement, and cause.
• In order for a force to qualify as having done work on an object, there must
be a displacement and the force must cause the displacement.
In each case
described here
• Examples where work is done…
there is a force
- A father pushing a grocery cart down the aisle of a grocery store exerted upon an
object to cause that
- A weightlifter lifting a barbell above his head
object to be
- A horse pulling a plough through the field
displaced.
Work
• Is work done in the following scenarios?
- A teacher applies a force to a wall and becomes exhausted
The wall is not displaced – A force must cause
displacement for work to be done
Work requires a
force to be
exerted upon an
object to cause
that object to be
displaced in the
direction of the
force.
- A falls off a table and free falls to the ground
Gravity (force) acts on the book causing it to be displaced downward
- A rocket accelerates through space
The force (expelled gases) push the rocket, causing displacement through space
Work
Calculating Work:
𝑊 = 𝐹𝑥
𝑊 − 𝑊𝑜𝑟𝑘 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝐹 − 𝐹𝑜𝑟𝑐𝑒, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑁𝑒𝑤𝑡𝑜𝑛𝑠 𝑁
𝑥 − 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑚𝑒𝑡𝑟𝑒𝑠 (𝑚)
Work
𝑊 = 𝐹𝑥
Eg1. A brick is lifted from the ground with a force of 100N, to a platform 2 metres
above. What work was done on the brick?
𝑊 = 𝐹𝑥
= 100 × 2
= 200 𝐽
Eg2. A wheelbarrow is pushed with a force of 400N, to a point 10 metres away.
What work was done on the wheelbarrow?
𝑊 = 𝐹𝑥
= 400 × 10
= 4000 𝐽
Work
𝑊 = 𝐹𝑥
eg3. A mum pushes a pram horizontally with a force of 200 N to move the pram a
distance of 8 metres. If the friction force opposing the motion is 120 N,
How much work is done on the pram by:
a) The force applied by the mum to push the pram?
b) The net force?
𝑊 = 𝐹𝑥
= 200 × 8
= 1600 𝐽
𝐹𝑛𝑒𝑡 = 200𝑁 − 120𝑁 = 80 𝑁
𝑊 = 𝐹𝑥
= 80 × 8
= 640 𝐽
c) The shopper to oppose the friction force?
𝑊 = 𝐹𝑥
= 120 × 8
= 960 𝐽
Work
Sometimes when a force acts, the displacement does not have the same
direction as the force. In these problems, we use:
𝑊 = 𝐹𝑥 cos 𝜃
eg. If the cart pictured was pulled with a 80 N force, and
the rope was inclined at a 40° angle, how much work was
done to pull the cart 10 metres?
𝑊 = 𝐹𝑥 cos 𝜃
= 80 × 10 × cos 40
= 800 cos 40
= 612.8 𝐽
Energy
All matter possesses energy
Energy comes in various forms…..
Nuclear
Energy
Electrical
Energy
Chemical
Potential
Energy
Kinetic Energy
Gravitational
Potential
Energy
Etc…..
Light Energy
Energy
• The law of conservation of Energy
Energy cannot be created or destroyed
• Energy can be stored, transferred to other matter,
or transformed from one form to another.
• Some energy transfers and transformations can be
seen, felt, heard, smelt, however many are not
observable and cannot be measured.
Kinetic Energy
Is the energy associated with the movement of an object, given by the
equation:
𝐸𝐾 =
1
𝑚𝑣 2
2
𝐸𝐾 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝑚 − 𝑚𝑎𝑠𝑠 𝑘𝑔
𝑣 − 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑚𝑠 −1 )
Kinetic Energy
𝐸𝐾 =
1
2
𝑚𝑣
2
A 70kg runner moves with a velocity of 6 m/s. What is his kinetic energy?
𝐸𝐾 =
1
× 70 × 62
2
1
× 70 × 36
2
=
= 1260 𝐽
A 1000 kg car moves with a velocity of 15 m/s. What is the kinetic energy of the car?
𝐸𝐾 =
1
× 1000 × 152
2
1
× 1000 × 225
2
=
= 112 500 𝐽
Kinetic Energy
The Kinetic Energy of an object can only change as a result of a non-zero net force
acting on it in the direction of the motion.
The change in Kinetic Energy of an object is equal to the work done on it by the net
force acting on it.
So we Kinetic Energy can be expressed as:
Change in Kinetic Energy = Work Done by the force acting on the object
∆𝐸𝐾 = 𝐹𝑁𝑒𝑡 𝑥
Now Do
Chapter 7
Questions 7, 8, 11, 12, 13
Potential Energy
Energy that is stored is called Potential Energy. These include: Chemical Potential
Energy, Gravitational Potential Energy, etc…
Our course covers Gravitational Potential Energy.
Gravitational potential energy is the energy stored in an object as a result of
its position and its attraction to the earth by the force of gravity.
So if something can fall down from somewhere, then it has Gravitational
Potential energy.
Gravitational Potential Energy
Is the energy associated with its position, given by the equation:
𝐸𝑔𝑝 = 𝑚𝑔ℎ
𝐸𝑔𝑝 − 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦, 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝐽𝑜𝑢𝑙𝑒𝑠 𝐽
𝑚 − 𝑚𝑎𝑠𝑠 𝑘𝑔
ℎ − ℎ𝑒𝑖𝑔ℎ𝑡 𝑓𝑟𝑜𝑚 𝑒𝑎𝑟𝑡ℎ 𝑜𝑟 𝑎 𝑠𝑢𝑟𝑓𝑎𝑐𝑒(𝑚)
Gravitational Potential Energy
𝐸𝑔𝑝 = 𝑚𝑔ℎ
A 40kg air conditioner is mounted to a wall 3 metres above the ground. What
gravitational potential energy does the air conditioner have?
𝐸𝑔𝑝 = 𝑚𝑔ℎ
= 40 × 10 × 3
= 1200 𝐽
A skater with mass 60kg stands at the top of a 4 metre skate ramp. What
gravitational potential energy does the skater have?
𝐸𝑔𝑝 = 𝑚𝑔ℎ
= 60 × 10 × 4
= 2400 𝐽
Conservation of Mechanical Energy
Kinetic (KE) and Gravitational Potential Energy (GPE) are referred to as Mechanical
Energy.
One type of energy can transform from Potential to Kinetic, or vice versa, if the object
with the energy, interacts with a force.
Eg. A tennis ball sits at a height of 2 metres.
• What type of mechanical energy does it possess?
• If its dropped to the ground, during this motion, its GPE is transformed throughout
the journey, from GPE to KE.
• At the bottom of the 2 metres, just before it hits the ground, all GPE is transformed
into KE.
Conservation of Mechanical Energy
Say a ball has a mass of 500g and is dropped from at height of 2 metres.
What total mechanical energy does the ball have at the top of the path, before
being dropped?
𝐸𝑔𝑝 = 𝑚𝑔ℎ
= 0.5 𝑘𝑔 × 10 𝑚/𝑠 2 × 2
= 10 𝐽
What type of energy the ball possess when it’s 1 metre from the ground?
GPE and KE.
What total energy does the ball have when it’s 1 metre from the ground?
Total Energy is still 10 𝐽 – But now this energy consists of BOTH KE and GPE.
𝐸𝑔𝑝 = 𝑚𝑔ℎ
The remaining 5 J is
= 0.5 × 10 × 1
Kinetic Energy
=5𝐽
Conservation of Mechanical Energy
How fast is the ball travelling at the 1 metre point?
𝐸𝐾 = 5 𝐽
1
2
𝐸𝐾 = 𝑚𝑣 2
1
2
5 = × 0.5 × 𝑣 2
𝑣2 =
5
0.25
= 20
𝑣 = 20 = 4.47 𝑚/𝑠
Energy Skate Park
Now Do
Chapter 7
Questions 14, 16, 19, 20, 23
Strain Potential Energy - Hooke’s Law
The potential energy stored in a spring is known as Strain Potential Energy.
The force needed to extend or compress a spring by some distance is proportional
to the distance in which it moves.
𝐹 ∝ ∆𝑥
Each spring has an individual stiffness associated with it, which is a measure of
how strong the spring is. This is called the spring constant represented by ‘ k ’
So, the Force required to extend or compress a spring can be found by:
𝐹 = 𝑘∆𝑥
F (Force applied) measured in N (Newtons)
k (Spring Constant) measured in N/m (Newtons per metre)
x (displacement of the spring) measured in m (metres)
Strain Potential Energy - Hooke’s Law
The Strain Potential Energy stored in a spring, whether compressed or extended,
can be found by calculating the area under a Force versus displacement graph for
the spring.
The spring constant k ,
can be found by
calculating the
gradient of the line.
Strain Potential Energy - Hooke’s Law
Eg. The graph below models a spring being stretched.
a) Find the Potential Energy gained by stretching the spring 25 cm
b) Calculate the spring constant.
a) 𝐸𝑃𝑂𝑇 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑎 𝐹 − 𝑥 𝑔𝑟𝑎𝑝ℎ
1
= 𝑏ℎ
2
1
2
= × 0.25𝑚 × 20𝑁
= 2.5 𝐽
b) 𝑘 = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
=
20−0
0.25−0
𝑦2 −𝑦1
𝑥2 −𝑥1
= 80 𝑁/𝑚
Strain Potential Energy - Hooke’s Law
We can also calculate the Strain Potential Energy stored in a spring using the
following equation:
1
𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = × 𝑘 × (∆𝑥)2
2
eg. A spring a stretched 15 cm. If the spring constant for the spring is 40 N/m, find:
a) The force applied to the spring
𝐹 = 𝑘∆𝑥
= 40 × 0.15
=6𝑁
b) The strain potential energy stored in
the extended spring
𝑆𝑃𝐸 =
1
𝑘(∆𝑥)2
2
1
× 40 × (0.15)2
2
=
= 0.45 J
Strain Potential Energy - Hooke’s Law
eg. A ball with mass 400g, on a pinball machine is released by extending & releasing a
spring. If the spring with a spring constant of 200 N/m is pulled back 15 cm, find:
a) The strain potential energy of the spring just before the ball is released.
𝑆𝑡𝑟𝑎𝑖𝑛 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 =
1
2
2
× 𝑘 × (∆𝑥) =
1
2
× 200 × 0.152 = 2.25 𝐽
b) The kinetic energy of the ball just after it has been released.
All SPE converted into KE, so 𝐸𝐾 = 𝑆𝑃𝐸 = 2.25 𝐽
c) The speed of the pinball just after it has been released.
1
2
1
2
𝐸𝐾 = 𝑚𝑣 2
→ 2.25 = × 0.4 × 𝑣 2
𝑣 2 = 11.25
→ 𝑣 = 50 = 3.35 𝑚/𝑠
Now Do
Chapter 7
Questions 18, 26, 27, 28