Find the net work done by friction on a box that moves in a complete circle of radius $1.82 \mathrm {~m}$ on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs $65.0 \mathrm {~N}$
A)190 J
B)0 J
C)1800 J
D)370 J

Question

Quizplus · Accepted Answer

The net work done by friction is calculated using the formula $ W = f_k \cdot d $, where $ f_k $ is the kinetic friction force and $ d $ is the distance traveled. The friction force $ f_k = \mu_k \cdot N = 0.25 \cdot 65.0 \mathrm{~N} = 16.25 \mathrm{~N} $. The distance $ d $ is the circumference of the circle, $ 2\pi \times 1.82 \mathrm{~m} \approx 11.43 \mathrm{~m} $. Thus, the work done by friction is $ 16.25 \mathrm{~N} \times 11.43 \mathrm{~m} \approx 190 \mathrm{~J} $.

Find the Net Work Done by Friction on a Box 1.82 m1.82 \mathrm {~m}1.82 m

Find the Net Work Done by Friction on a Box $1.82 \mathrm {~m}$