Question

Consider the system shown in figure. Coefficient of friction between the block and table is $\mu =0.5$. The system is released from rest. Find the work done by friction, when the speed of block is $10m/s(m=1kg)$

• A. $-10J$
• B. $-20J$
• C. $-30J$
• D. $-50J$

Answer 1

The correct option is D $-50J$

REF. Image

Let the displacement of block be

s. Then,

work force = $-f_s.s$

consider the FBD

$2Mg-T=2mg$ ____ (i)

$T-\mu mg=ma$

$\Rightarrow T-\frac{mg}{2}=ma$ ___ (ii)

Adding (i) and (ii), we get

$3ma=\frac{3mg}{2}$

$a=g/2$

To find displacement at v = 10 m/s

Applying $3^{rd}$ $e{q}^n$ of motion,

$v^2-u^2=2as$

$\Rightarrow \left(100\right)-0=2\times \frac{g}{2}s$

$\Rightarrow s=\frac{100}{g}=\frac{100}{10}=10m$

work-done by friction = $(-\frac{mg}{2})s$

$=-\frac{10}{2}\times 10=-50J$

Option - D is correct.