Energy and work physics problem

Question

During the first snowfall of the year a 23-kg physics textbook slides 18 m on level ground before coming to rest. If 430 J of work are done by friction, calculate the coefficient of friction between the book and the ground.

Answer 1

Draw a free-body diagram:

F_F is the force of friction (which results in the work done by friction); F_G is the force of gravity (and we know that F_G = mg); and F_N is the normal force. Since there is no movement in the y-direction, then ∑F_y = 0 so F_G = F_N

mg = F_N

(23)(9.81) = F_N

F_N = 226 N

We know that F_F = μF_N where μ is the coefficient of friction (note that this is the kinetic coefficient of friction).

So F_F = 226μ

In considering the work done by friction, W = F_F(d) and since W = 430 J and d = 18 m then 430 = 18F_F

We can now plug F_F = 226μ into this work equation to get:

430 = 18(226μ) and we can divide to get μ = 430/(18•226) = 0.11