Question

(a) The Eskimo pushes the same 50.0-kg sled over level ground with a force of 1.75 102 N exerted horizontally, moving it a distance of 6.00 m over new terrain. If the net work done on the sled is 1.50 102 J, find the coefficient of kinetic friction.

Answer 1

Answer: 0.306

Explanation:

from the question we are given the following

mass of sled (m) = 50 kg

force (f) = 1.75 x 10^2 N = 175 N

distance (s) = 6 m

net work done on the sled = 1.50 x 10 ^2 N = 150 N

acceleration due to gravity (g)  = 9.8 m/s^2

coefficient of friction = μ

lets first calculate the frictional force (ff)

ff =  μ x m x g = μ  x 50 x 9.8 = 490 μ

work done on the slide by the applied force (W1)=  f x s = 175 x 6 = 1050 j

work done on the slide by frictional force (W2) = ff x s = 490 μ x 6 = 2940μ j

now the net work done is the work done by the frictional force subtracted from the work done by the applied force

net work done = W1 - W2

150 = 1050 -  2940μ

2940μ = 1050 - 150

μ = 900 / 2940

μ = 0.306