1. Total Energy before = Total Energy After
Conservation of Energy
Total Energy before = Total Energy After
Comes from = Goes to
Assets = Expenditures
Fs + mgh + 1/2mv2 + 1/2kx2 = Fs + mgh + 1/2mv2 + 1/2kx2

2. Fs + mgh + 1/2mv2 + 1/2kx2 = Fs + mgh + 1/2mv2 + 1/2kx2
A 250 kg cart going 4.5 m/s rolls down a 1.75 m tall hill. What is the velocity of the cart at the bottom?
v = 4.5 m/s
250 kg
1.75 m
Fs + mgh + 1/2mv2 + 1/2kx2 = Fs + mgh + 1/2mv2 + 1/2kx2
7.4 m/s

3. Example 1
An 890 kg cart rolling 6.2 m/s along a level surface hits a 3.6 m long puddle that exerts 3200 N of average retarding force. What is the cart’s velocity after this?
v = 6.2 m/s
890 kg
3.6 m
(Puddle - Exerts 3200 N of retarding force)
3.5 m/s

4. Example 2
A 350 kg cart is going 4.6 m/s. For what distance must a person exert a forward force of 53 N so that when the cart gets to the top of a 1.8 m tall hill it is going 2.4 m/s?
u = 4.6 m/s
350 kg
1.8 m
66 m

5. Example 3
A kg pine cone falls 45.0 m from a tree. It is going 22.0 m/s when it strikes the ground. What is the average force of air friction that acts on the pine cone as it falls?
0.550 N

6. (15 kg)(9.81 N/kg)(2.15 m) = 1/2(15 kg)v2
u = 0
What speed at the bottom?
h = 2.15 m
(15 kg)(9.81 N/kg)(2.15 m) = 1/2(15 kg)v2
v = …
6.5 m/s

7. (15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2
u = 5.8 m/s
What speed at the bottom?
h = 2.15 m
(15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2
v = …
8.7 m/s

8. What speed at the top? 15 kg h = 4.25 m h = 2.15 m u = 8.6 m/s
(15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(8.6 m/s)2 =(15 kg)(9.81 N/kg)(4.25 m) 1/2(15 kg)v2
v = 5.723…
5.7 m/s

9. v = 3.68 m/s
2.34 kg
The hammer pushes in the nail 3.50 mm. ( m). What force did it exert on the nail
1/2(2.34 kg)(3.68 m/s)2 = F(.0035 m)
4530 N

10. (125 kg)(9.81 N/kg)(18.0 m + 1.50 m) = F(1.50 m)
A 125 kg experiment falls 18.0 m and has its velocity arrested in 1.50 m by an air bag. What force does the air bag exert on the experiment in stopping the experiment?
18 m
(125 kg)(9.81 N/kg)(18.0 m m) = F(1.50 m)
F = 15,941.25
15,900 N

11. (450 kg)(9.81 N/kg)(1.75 m) + 1/2(450 kg)(5.8 m/s)2 = 1/2(15000 N/m)x2
What distance will it compress the N/m spring in stopping the cart?
450 kg
u = 5.8 m/s
h = 1.75 m
(450 kg)(9.81 N/kg)(1.75 m) + 1/2(450 kg)(5.8 m/s)2 = 1/2(15000 N/m)x2
x = 1.428…
1.4 m

12. What must be the spring constant to give the 115 g marble a velocity of 2.13 m/s on top of the hill if the spring is compressed 3.15 cm?
h = .452 m
1/2k(.0315 m)2 = 1/2(.115 kg)(2.13 m/s)2 + (.115 kg)(9.81 N/kg)(.452 m)
k = …
1550 N/m