Dynamics: Conservation of energy
Work Energy theorem
Problems

2. Problem 2:

An object of mass 1 kg travelling at 5.0 m/s enters a region of ice where the coefficient of kinetic friction is .10. Use the work energy theorem to find the distance the object travels before coming to rest.

Solution:

The work energy theorem gives W = Δ KE .
We have W = - f_kd = - μ_k Nd = - μ_k mg d and ΔKE = (1/2)mvf² (=0) - (1/2)mvi² = - (1/2)mvi2 .

Combining,
- μ_k mg d = - (1/2)mvi2

d = vi² /2μ_kg = 13 m.

3. Problem 3:

A 30 kg child enters the final section of a waterslide travelling at 2.0 m/s. The final section is 5.0 m long and has a vertical drop of 3.0 m. The force of friction opposing the child's motion is 50 N. Find a) the loss of potential energy, b) the work done by friction in the final section and c) the child's velocity at the end of the section (using energy considerations).

Solution:
a)
ΔPE = mg(hf - hi)
= 30(9.8)(0 - 3) = - 882 J

b)
W = - f_kx = - 50(5) = - 250 J

c)
Wnc = ΔKE + ΔPE

- 250 = (1/2)(30)(vf2) - (1/2)(30)(2.0)2 - 882

vf = 6.8 m/s

4. Problem 4

A 2.0 kg wood block is on a level board and held against a spring of spring constant k=100 N/m which has been compressed .1 m. The block is released and pushed horizontally across the board. The coefficient of friction between the block and the board is $\mu_{k}^{}$ = .20. Find a) the velocity of the block just as it leaves the spring and b) the distance the block travels after it leaves the spring.

Solution:

a)
The work energy theorem gives:

Wnc = ΔKE + ΔPE

- f_kx = (1/2) mvf2 - 0) + (0 - (1/2))kx 2
- mu_k mgx = (1/2) mvf2 - (1/2)kx 2

vf2 = ...

vf = 0.33 m/s.

b)
The work energy theorem gives, - mu_k mgd = 0 - (1/2)mvi2

d = ...

d = 0.028 m.

5. Problem 5

A man pushes a 100 kg box across a level floor at a constant speed of 2.0 m/s for 10 s. If the coefficient of friction between the box and the floor is μ_k = 0.20 , find the average power output by the man.

Solution:

Since the acceleration of the box is zero the force exerted by the man is obtained from F_man - f_k = 0 . Then

F_man = f_k = μ_k mg . Then

P = W/Δt = fv = μ_k mgv = 0.2(100)(9.8)(2) = 392 W.