A box weighing $2000 \mathrm{~N}$ is to be slowly slid through $20 \mathrm{~m}$ on a straight track having friction coefficient $0.2$ with the box. (a) Find the work done by the person pulling the box with a chain at an angle 9 with the horizontal. (b) Find the work when the person has chosen a value of 0 which ensures him the minimum magnitude of the force.
(a) $R+P \sin \theta=2000 \ldots \ldots .1$
$P \cos \theta-0.2 R=0$
Solving 1 and 2 equations we get,
$P=\frac{400}{(\cos \theta+0.2 \sin \theta]}$
and work done $\mathrm{W}=\mathrm{PS} \cos \theta$
$P=\frac{400}{(\cos \theta+0.2 \sin \theta)}$
and work done $\mathrm{W}=\mathrm{PS} \cos \theta$
$W=\frac{40000}{(5+\tan \theta)}$
(b) For minimum force,
$\frac{d}{d \theta}[\cos \theta+0.2 \sin \theta]=0$ (from equation 1)
Also, $W=40000 / 5+\tan \theta=40000 / 5+0.2$