Question

The coefficient of kinetic friction between a $20 \,kg$ box and the floor is $0.40$. How much work does a pulling force do on the box in pulling it $8.0\, m$ 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, $N=m g-F \sin 37^{\circ}$,
So that, $F=\frac{\mu mg }{\left(\cos 37^{\circ}+\mu \sin 37^{\circ}\right)}$
Here, $\mu=0.40$ and $m=20 \,kg$
$\therefore F=75.4 \,N$
Hence, $W=\left(75.4 \cos 37^{\circ}\right)(8.0)=482\,J$