Question

A block of mass $1\text{kg}$ is at rest relative to a smooth wedge moving leftwards with constant acceleration $a=5{\text{m/ s}}^2$. Let $N$ be the normal reaction between the block and the wedge. Then
$(g=10{\text{m/s}}^2$)

• A. $N=5\sqrt{5}\text{N}$
• B. $N=15\text{N}$
• C. $\tan \theta =\frac{1}{2}$
• D. $\tan \theta =2$

Answer 1

Answer: (A), (C)

The correct options are
A $N=5\sqrt{5}\text{N}$
C $\tan \theta =\frac{1}{2}$

Given block of mass $1\text{kg}$ at rest on the smooth inclined plane.
Wedge has an acceleration of $5{\text{m/s}}^2$ so a pseudo-force of $5\text{N}$ will act on block
And it’s direction will be Negative to the direction of wedge.
So, the block has these three forces acting on it
$mg=10\text{N}$ (downward), Normal (at an angle $\theta$ from vertical) and $5\text{N}$ rightward
The normal force will be the resultant of $mg$ and pseudo-force
$=\sqrt{{10}^2+5^2}$
$=\sqrt{125}=5\sqrt{5}$
We have $\tan \theta =\frac{ma}{mg}$
$=\frac{5}{10}=\frac{1}{2}$
$N\sin \theta =ma$
$N\cos \theta =mg$
$N=m\sqrt{a^2+g^2}$ and $\tan \theta =\frac{a}{g}$