Question

A 15.2 kg box is slid along the ground with a coefficient of kinetic friction of 0.569.

a. If the box moves 8.04 m, what is the amount of work done by friction on the box?

b. If the box moves 6.12 m and has an acceleration of 0.972 m/s^2. what is the work done on the box by the applied force?

Answer 1

Mass of box, $m=15.2\space kg$

Coefficient of friction, $\mu_k=0.569$

(a) Displacement, $s=8.04\space m$

Frictional Force, $F_\mu=\mu mg=0.569\times15.2\times9.8=84.75\space N$

Work done by frictional force $=F_\mu\times s\times cos\theta=84.75\times8.04\times cos180\degree$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=-681.45 \space J$

(b) Acceleration, $a=0.972\space m/s^2$

Force Applied, $F=ma=14.77\space N$

Displacement, $s=6.12\space m$

Work Done $=F\times s\times cos\theta=14.77\times6.12\times cos90\degree=90.41\space J$