Question

A block of mass $1\text{kg}$ is at rest relative to a smooth wedge moving leftwards with constant acceleration $a=5{\text{ms}}^{-2}.$ Let $N$ be the normal reaction between the block and the wedge. Take $g=10{\text{ms}}^{-2}$. Choose the correct options from the choices given below:

• 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 $\tan \theta =\frac{1}{2}$


Since block is at rest w.r.t. wedge, the system of (block + wedge) is moving together with acceleration $a$ leftwards.

FBD of Block:


Solving from ground frame:

Applying equilibrium condition in vertical direction

$N\cos \theta =mg....\left(i\right)$

Applying Newton's 2nd law in direction of acceleration:

$N\sin \theta =ma....\left(ii\right)$

Dividing Eq. $\left(ii\right)$ and $\left(i\right)$ :
$\tan \theta =\frac{a}{g}=\frac{5}{10}=\frac{1}{2}$


Putting the value of $\sin \theta$ from triangle shown in figure, in Eq. $\left(ii\right)$:

$N\sin \theta =1\times 5$
$\Rightarrow N\times \frac{1}{\sqrt{5}}=5$
$\therefore N=5\sqrt{5}\text{N}$

Hence, options A and C are correct.