Question

A plank $P$ and block $Q$ are arranged as shown on a smooth table top. They are given velocities $3m/s$ and $6m/s$ respectively. The length of plank is $1m$ and block is of negligible size. After some time when the block has reached the other end of plank it stips slipping on plank. The velocity of plank then is (coefficient of friction between plank and block is $0.3$ and mass of plank is double of block).

• A. $4m/s$
• B. $5m/s$
• C. $4.5m/s$
• D. Zero

Answer 1

The correct option is A $4m/s$

frictional force on block Q is;

$f=\mu \left(N\right)=0.3mg$

Work done (W) by friction, when Q reach the other end of plank P is;

$W=lf$

where,

$l-$ length of plank

$f-$ frictional force

$W=1\times \left(0.3mg\right)=0.3mg$

Applying energy conservation;

$\frac{1}{2}m{v}_1^2$ $+\frac{1}{2}\left(2m\right)v_2^2$ $=\frac{1}{2}(m+2m)v^2+W$

Mass of plank is double of block Q;

v- combined velocity

When Q at other end of plank;

$m(6{)}^2+2m(3{)}^2=m(3v{)}^2+2\times 0.3\times g$

$36+18=3v^2+6$

$3v^2=48$

$v^2=16$

$v=4m/s$