Q. 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 :

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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 $4 \, N$.

Solution:

$N=m g \cos \theta+m a \sin \theta$
$N=1 \times 10 \times \frac{4}{5}+1 \times 5 \times \frac{3}{5}$
$N=8+3=11 \,N$
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,
$a=\frac{m g \sin \theta-m a \cos \theta}{m}$
$a=\frac{6-4}{1}=2 \,m / s ^{2}$

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