Question

The coefficient of kinetic friction between a $20kg$ box and the floor is $0.40$. How much work does a pulling force do on the box in pulling it $8.0m$ across the floor at constant speed? The pulling force is directed $37^{\circ }$ above the horizontal

• A. 343 J
• B. 482 J
• C. 14.4 J
• D. None of these

Answer 1

Answer: (B)

The work done by the force is $F\cos 37^{\circ }$,
where $F\cos 37^{\circ }=f=\mu N$

In this case, $N=mg-F\sin 37^{\circ }$,
So that, $F=\frac{\mu mg}{\left(\cos 37^{\circ }+\mu \sin 37^{\circ }\right)}$
Here, $\mu =0.40$ and $m=20kg$
$\therefore F=75.4N$
Hence, $W=\left(75.4\cos 37^{\circ }\right)(8.0)=482J$