Question

A block of mass $m$ is pulled on a rough horizontal surface which has a friction coefficient $\mu$. A horizontal force $F$ is applied which is capable of moving the body uniformly with speed $v$. Find the work done on the block in time $t$ by (a) weight of the block (b) Normal reaction by surface on the block (c) friction (d) $F$

Answer 1

$\text{Step 1: F.B.D of block [Refer Fig.]}$

As the velocity is constant, so acceleration $a=0$

Since slipping is there so kinetic friction will act i.e. $f=\mu N=\mu mg$

From Newton's second law:

$F_{net}=ma$

$F-f=m\times 0$

So, $F=f=$ $\mu mg$ $....\left(1\right)$

$\text{Step 2: Work done by weight(mg) of block}$

Angle between $\left(mg\right)$ and direction of motion is ${90}^{\circ }$

So, $W=F.d\cos \theta$

$=\left(mg\right).d\cos {90}^o$

$W=0$ ...............$\left(a\right)$

$\text{Step 3: Work done by normal reaction}$

Angle between normal and direction of motion is ${90}^{\circ }$

So, $W=N.d\cos {90}^o$

$W=0$ ...............$\left(b\right)$

$\text{Step 4: Workdone by friction force}$

Let displacement by body is $S$

So, $S=vt$

Now,

Work done $W=f.S\cos \theta$

$=\mu N.S\cos \left({180}^{\circ }\right)$ $\theta ={180}^{\circ }$

$\therefore W=\mu mg\left(vt\right)(-1)$ $=-\mu mgvt$ ............$\left(c\right)$

$\text{Step 5: Work done by F}$
$W_F=F.S\cos \theta$ $=F.S\cos 0^o$ $\theta =0^{\circ }$

$W_F=F\left(vt\right)\left(1\right)$ $=Fvt$

Work done by $\left(F\right)=Fvt$ $=\mu mgvt$ ...............$\left(d\right)$ (Using Equation $1$)

So, Comparing answers of (c) and (d), Work done by $F=-ve$ of Work done of friction