Find the how much mechanical energy was converted to internal energy in the system box-surface-Earth when a box 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 mechanical energy converted to internal energy due to friction can be calculated using the work done by friction force. The work done by friction is given by $W = f \cdot d$, where $f$ is the frictional force and $d$ is the distance. The frictional force $f$ can be found using $f = \mu \cdot N$, where $\mu$ is the coefficient of kinetic friction and $N$ is the normal force. In this case, the normal force $N$ is equal to the weight of the box, which is $65.0 \mathrm{~N}$, and $\mu = 0.25$. The distance $d$ the box moves is the circumference of the circle, which is $2\pi r$, where $r = 1.82 \mathrm{~m}$. Therefore, the work done by friction (energy converted to internal energy) is $W = 0.25 \cdot 65.0 \cdot 2\pi \cdot 1.82 \approx 190 \mathrm{~J}$.

Find the How Much Mechanical Energy Was Converted to Internal 1.82 m1.82 \mathrm {~m}1.82 m

Find the How Much Mechanical Energy Was Converted to Internal $1.82 \mathrm {~m}$