JEE Main 2026 — Laws of Motion Question with Solution

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):
$A=\frac{f}{M+m}=\frac{24}{10+2}=2{\text{ m/s}}^2$

Working in the non-inertial frame of the wedge, the block experiences a pseudo force $mA$ directed opposite to the wedge's acceleration. The components along the incline are:
Component of gravity down the incline: $g\sin 37^{\circ }$
Component of pseudo force up the incline: $A\cos 37^{\circ }$

Net relative acceleration of the block down the incline:
$a_{\text{rel}}=g\sin 37^{\circ }-A\cos 37^{\circ }$

Using $\sin 37^{\circ }=0.6$ and $\cos 37^{\circ }=0.8$:
$a_{\text{rel}}=10(0.6)-2(0.8)=6-1.6=4.4{\text{ m/s}}^2$

Time taken to slide a distance of $8.8\text{ m}$ along the incline, starting from rest:
$s=\frac{1}{2}{a}_{\text{rel}}{t}^2$

$8.8=\frac{1}{2}(4.4)t^2$

$8.8=2.2t^2$

$t^2=4\Rightarrow t=2\text{ s}$

Hence, the required time is $2\text{ s}$.