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

Answer 1

Answer: (A), (C)

$\tan \theta =\frac{1}{2}$

Solving from wedge's frame

$\therefore N\sin \theta =ma\dots \dots \left(i\right)$
and $N\cos \theta =mg\dots \dots \left(ii\right)$
$\tan \theta =\frac{a}{g}=\frac{5}{10}=\frac{1}{2}$

$\Rightarrow N=\frac{ma}{\sin \theta }=\frac{1\left(5\right)}{\frac{1}{\sqrt{5}}}$
$\Rightarrow N=5\sqrt{5}N$