Question

Figure shows a wedge of mass $2\text{kg}$ resting on a frictionless floor. A block of mass $1\text{kg}$ is kept on the frictionless wedge and the wedge is given an acceleration of $5{\text{m/s}}^2$ towards right. Then select the correct statement. (Assume $g=10m/s^2$)

• A. the block will remain stationary w.r.t wedge
• B. the block will have an acceleration of $1{\text{m/s}}^2$ w.r.t wedge
• C. the normal reaction on the block is $11\text{N}$
• D. None of the above

Answer 1

The correct option is C the normal reaction on the block is $11\text{N}$

In the Non-inertial frame of wedge and block, we have
F.B.D. of $1\text{kg}$ block.


After resolving the forces,


a) Resolving forces along the wedge surface,
$F=\text{mg}\sin \theta -ma\cos \theta$
$\Rightarrow F=mg\sin {37}^{\circ }-ma\cos {37}^{\circ }$
$\Rightarrow F=1\times 10\times \frac{3}{5}-1\times 5\times \frac{4}{5}$
$\Rightarrow F=2\text{N}$
Hence, $1\text{kg}$ block will not be stationary because of $2\text{N}$ force.
b) $F=ma$ (block is allowed to move only along the wedge surface)
We got $F=2\text{N}$ along the wedge surface, so acceleration of the block w.r.t wedge
$F=ma\Rightarrow 2=1\times a$
$\Rightarrow a=2{\text{m/s}}^2$

c) $N=mg\cos \theta +ma\sin \theta$
$\Rightarrow N=mg\cos {37}^{\circ }+ma\sin {37}^{\circ }$
$\Rightarrow N=1\times 10\times \frac{4}{5}+1\times 5\times \frac{3}{5}$
$\Rightarrow N=11\text{N}$
$\therefore$ Normal force $=11\text{N}$
Hence, option (c) is correct.