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
