Q. 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
Explanation
(b) The work done by the force is $F$ cos $37{^\circ}$,
Where $F\cos 37{^\circ} = f = \mu N$
[image:5_image181.emf]
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 $m = 20$ kg
$\therefore\ F = 75.4$ N
Hence, $W = \left( 75.4\cos{37{^\circ}} \right)(8$.0$) = 482\ $J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.