Solve the following :


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, $W=40000 / 5+\tan \theta=40000 / 5+0.2$


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