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 done by the person pulling the box with a chain at an angle $\theta$ with the horizontal.

• A.
• B. 8000 J
• C.
• D. None of these

Answer 1

The correct option is C

$w=2000N$

As the body is being moved slowly we can assume it is moving with constant speed.

$\sum F_y=0$

$N+Fsin\theta =mg$

$\Rightarrow N+Fsin\theta =2000N$

$\Rightarrow N=2000-Fsin\theta$

$\sum F_x=0$

$\Rightarrow f=Fcos\theta$

$\Rightarrow \mu N=Fcos\theta$

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

$\Rightarrow \frac{1}{5}(2000-Fsin\theta )=Fcos\theta$

$\Rightarrow 2000-Fsin\theta =5\left(Fcos\theta \right)$

$\Rightarrow 2000=5Fcos\theta +Fsin\theta$

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

$\Rightarrow F=\frac{2000}{5cos\theta +sin\theta }$

$\therefore workdone=F.scos\theta$

= $\frac{2000}{5cos\theta +sin\theta }\left(cos\theta \right)\times 20$

=$\frac{4000}{5+tan\theta }J$