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 = \mu_k \cdot N $ is the frictional force, $ \mu_k = 0.25 $ is the coefficient of kinetic friction, $ N = 65.0 \mathrm{~N} $ is the normal force, and $ d = 2\pi \times 1.82 \mathrm{~m} $ is the distance traveled in a circle. Calculating gives $ f_k = 0.25 \times 65.0 = 16.25 \mathrm{~N} $ and $ d = 2\pi \times 1.82 \approx 11.43 \mathrm{~m} $. Thus, $ W = 16.25 \times 11.43 \approx 185.86 \mathrm{~J} $, which rounds to 190 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 $1.82 \mathrm {~m}$