Question

A box weighing $2000N$ is to be slowly slid through $20m$ on a straight track having friction coefficient $0.2$ with the box. Find the work when the person has chosen a value of $\theta$ which ensures him the minimum magnitude of the force.

Answer 1

Here weight of the box$,mg=2000N$

$\mu =0.2$

distance $s=20m.$

.

so if the pull is applied by the person is F .

Then the Normal force, $N+F.sin\theta =W=2000N$

$N=2000-Fsin\theta$

Force of friction is $=\mu N$

$=\mu (2000-Fsin\theta )$

$=0.2(2000-Fsin\theta )$

$=400-0.2Fsin\theta$ .

According to the given conditions driving force is equal to the frictional force.

so, $F.cos\theta =400-0.2F.sin\theta$

$F(cos\theta +0.2sin\theta )=400$

multiply both sides by$5$ then,

$F(5cos\theta +sin\theta )=2000$

$F=2000/\left[\right(5cos\theta +sin\theta \left)\right]$

Now for work done

work done by the person pulling the box is

$=Fcos\theta \times 20J$

$=2000/\left[\right(5cos\theta +sin\theta \left)\right]cos\theta \times 20J$

$W=40000/(5+tan\theta )J$. -------(i)

Hence the work done by the person is$40000/(5+tan\theta )J.$

For minimizing the $F,df/d\theta =0$

$-2000(-5sin\theta +cos\theta )/(5cos\theta +sin\theta )²=0$

hence,$cos\theta =5sin\theta$

$sin\theta /cos\theta =1/5$

$tan\theta =1/5$

so by putting this tanθ value in equation (i), work done expression.

$W=40000/(5+1/5)J$

$W=7692J.$