Question

Figure-$2.162$ shows a wedge of mass $2\, kg$ resting on a frictionless floor. A block of mass $1\, kg$ is kept on the wedge and the wedge is given an acceleration of $5\, m / \sec ^{2}$ towards right. Then :

• A. Block will remain stationary w.r.t. wedge.
• B. The block will have an acceleration of $1\, m / \sec ^{2}$ w.r.t. the wedge.
• C. Normal reaction on the block is $11\, N$.
• D. Net force acting on the wedge is $4 \, N$.

Answer 1

Answer: (C)

$N=m g \cos \theta+m a \sin \theta$
$N=1 \times 10 \times \frac{4}{5}+1 \times 5 \times \frac{3}{5}$
$N=8+3=11 \,N$
As we have $m g \sin \theta>m a \cos \theta$,
The block cannot remain stationary w.r.t. wedge.
Acceleration of block w.r.t. wedge,
$a=\frac{m g \sin \theta-m a \cos \theta}{m}$
$a=\frac{6-4}{1}=2 \,m / s ^{2}$