Work-Energy Theorem
Unit 5: Energy and Work
Work, PE and KE
Work-Energy Theorem
Work-Energy Theorem
W = KEf – KEi
Fd = ½mvf2 – ½mvi2
W = ΔKE
Sample Problem 1
A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant 45 N force. How far must the student be pushed, starting from rest, so that her final KE is 352 J?
F = 45N
KEi = 0 J
KEf = 352 J
d = ?
W = ΔKE
Fd = KEf – KEi
45d = 352 – 0
d = 7.82 m
Sample Problem 2
A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 4.5 m starting from rest, its final speed is 6 m/s. Find the net force on the bobsled.
m = 75 kg
F = ?
KEi = 0 J
d = 4.5 m
vf = 6 m/s
W = ΔKE
Fd = KEf – KEi
Fd = ½ mvf2 – KEi
4.5F = ½(75)(62) – 0
4.5F = 1350
F = 300 N
Sample Problem 3
A 2.5 kg book is moving at 7.3 m/s initially. The book stops in 3 meters. Calculate the frictional force on the book.
m = 2.5 kg
F = ?
KEf = 0 J
d = 3 m
vi = 7.3 m/s
W = ΔKE
Fd = KEf – KEi
Fd = KEf – ½ mvi2
3F = 0 – ½(2.5)(7.32)
3F = -66.6
Ff = -22.2 N
Why is it negative??
Consider the Situation…
You are on a roller coaster that starts 40 meters high in a cart that has a total mass of 250 kg.
What is the TME at the top of the hill?
TME = KE + PE 🡪 TME = 0 + mgh 🡪 TME = (250)(9.8)(40) 🡪 TME = 98,000 J
Eventually the roller coaster stops on level ground. According to the Law of Conservation of energy, this should not happen. Your TME at the bottom should be equal to the TME at the top.
Why aren’t they equal?
Consider the Situation…
When work is applied, the system is no longer closed. By friction acting on the roller coaster causing it to stop, the energy is no longer conserved within the system.
Work-Energy Bar Charts
Final energy = 0 J because it is on the ground at rest.
Work-Energy Bar Charts
Final energy = KE + W
Lesson Check 5.4