Question

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

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

Answer 1

Answer: (A), (C)

$\tan \theta =\frac{1}{2}$
$N=5\sqrt{5}N$

Given block of mass $1kg$ at rest on the smooth inclined plane

Wedge has an acceleration of $5m/s^2$ so a pseudo-force of $5N$ will act on block

And its direction will be Negative to the direction of $a$

So the block has these three forces acting on it

$mg=10N$ (downward) ,Normal (at an angle $\theta$ from vertical) and $5N$ downward

The normal force will be the resultant of $mg$ and pseudo-force

= $\sqrt{10^2+5^2}$

= $\sqrt{125}$

= $5\sqrt{5}$

We have $\tan \theta =ma/mg$

= $5/10$

= $1/2$