1. Work and Energy Practice

2. Problem #1
A ball is dropped from rest at a height of 5.0 m above the ground. Ignoring air resistance, what is the speed of the ball just before it reaches the ground?

3. Gravity is the only force acting on the ball so mechanical energy is conserved.

4. Problem #2
A 2.0 kg box is released from rest at the top of a frictionless ramp. The ramp makes an angle of 30° to the horizontal and the box is 5.0 m above the ground at the top of the ramp. What is the speed of the box when it reaches the bottom of the ramp?

5. Frictionless, so ΔE=0
2.0 kg
5.0 m
30°

6. Problem #3
A frictionless ramp is placed on a table that is 1.5 m high. The ramp makes a 20 degree angle to the horizontal and the vertical height at the end of the ramp is 0.5 m. If the box is released from rest, what is the speed of the box right before it hits the ground? (Ignore air resistance)

7. Frictionless, so ΔE=0

8. Problem #4
A 20.0 N box is pushed along the floor by a force of 5.0 N that applied to the box at an angle of 40° above the horizontal. The coefficient of kinetic friction between the box and the floor is The box starts at position x = 0 m with a speed of 2.0 m/s. What is the speed of the box after it reaches position x = 4.0 m?

11. Problem #5
A 50 kg diver steps off a 10 m high diving board and drops straight down into the water. If the diver comes to rest 5 m below the surface of the water, determine the average resistance force exerted on the diver by the water.

12. Work required to stop
the diver
W = Fdcosθ
W = (F)(5)(cos180)
W = -5F
5 m

13. Set Ug =0 at bottom, K= 0 at bottom
W = ΔE
Set Ug =0 at bottom, K= 0 at bottom
ΔE = 0 - U at top of tower
Ug = (50)(9.8)(15) = 7350 J
ΔE = -7350
W = -7350
-5F = -7350
F = 1470 N

14. Problem #6
In a circus performance, a monkey on a sled is given an initial speed of 4 m/s up a 25º incline. The combined mass of the monkey and the sled is 20 kg, and the coefficient of kinetic friction between the sled and the incline is How far up the incline does the sled move?

15. mgsinq
mgcosq q q

16. vxo = 4 m/s
m = 20 kg
μ= 0.20 d
h = d sin 25 o 25

17. WFf =(0.2)(20)(9.8)(cos25)(d)(-1)= -35.5d
W = ΔE = Ug - K
Ug = (20)(9.8)(dsin25) = 82.8 d
K = ½ (20)(4)2 = 160J
WFf = Ff(d)(cos180)
Ff = µmgcosθ
WFf =(0.2)(20)(9.8)(cos25)(d)(-1)= -35.5d

18. W=ΔE
WFf =Ug – K
-35.5d =82.8d – 160
d = -160
d = 1.4 m