Question

A block is placed on an inclined plane moving horizontally towards right with an acceleration $a=g$. Take $g=10{\text{ms}}^{-2}$ .The length of plane $\text{AC}=1\text{m}$. Friction is absent everywhere. Time taken by the block to reach from $\text{C}$ to $\text{A}$ is

• A. $0.5\text{s}$
• B. $1\text{s}$
• C. $\sqrt{2}\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is B $1\text{s}$


Concept involved is application of pseudoforce, because the inclined plane is accelerating.
So,we will solve from frame of inclined plane therefore applying an additional force $F_{\text{pseudo}}=m_{\text{block}}\times a_{\text{inclined plane}}=m\times a=ma$

$\because$Block is moving up the inclined plane. Applying newton's $2^{\text{nd}}$ law along this direction gives,

$ma\cos \theta -mg\sin \theta =m{a}_r$
$a_r=g(\frac{4}{5}-\frac{3}{5})=\frac{g}{5}=2{\text{ms}}^{-2}$
Once we get the acceleration of block w.r.t. inclined plane, we can use kinematic equation of motion & replacing all parameters with relative parameters.

$S_r=u_r+\frac{1}{2}{a}_r{t}^2$
where, $S_r=\text{AC}=1\text{m}$ i.e relative displacement w.r.t inclined plane is $1\text{m}$, $u_r=0$
$\therefore$$S_r=\frac{1}{2}{a}_r{t}^2\Rightarrow t=\sqrt{\frac{2\times 1}{2}}=1\text{s}$