Question 60: Laws of Motion

Question

A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^{\circ}$ with horizontal. A block $X$ with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At $t=0$ wedge ( $Y$ ) is pulled toward right with constant force $(f)$ of 24 N . Taking the block $X$ at rest at $t=0$, the time taken by it to slide down 8.8 m on the slope, while $Y$ is on the move, is $\_\_\_\_$ s.
$\left(ight.$ take $	an \left(37^{\circ}ight)=3 / 4$ and $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2ight)$

Accepted Answer

The length of the incline (hypotenuse) is given by :
$$ \cos 	heta=\frac{L}{s} \Rightarrow s=\frac{L}{\cos 	heta} $$
Since the wedge is accelerating to the left with $\mathrm{a}_0$, we must work in the non-inertial frame of the wedge. This causes a pseudo-force ( $\mathrm{ma}_0$ ) acting on the block to the right.
We resolve all forces along the surface of the incline :
The net acceleration relative to the incline $\left(\mathrm{a}_{\mathrm{rel}}ight)$ is :
$$ \mathrm{a}_{\mathrm{rel}}=\frac{\mathrm{mg} \sin 	heta-\mathrm{ma}_0 \cos 	heta}{\mathrm{~m}}=\mathrm{g} \sin 	heta-\mathrm{a}_0 \cos 	heta $$
Using the kinematic equation for distance s starting from rest $(\mathrm{u}=0)$ :
$$ \mathrm{s}=\frac{1}{2} \mathrm{a}_{\mathrm{rel}} \mathrm{t}^2 . $$
Putting $\mathrm{s}=\mathrm{L} \cos 	heta$ and $\mathrm{a}_{	ext {rel }}=\mathrm{g} \sin 	heta-\mathrm{a}_0 \cos 	heta:$
$$ \frac{L}{\cos 	heta}=\frac{1}{2}\left(g \sin 	heta-a_0 \cos 	hetaight) t^2 $$
$$ \mathrm{t}^2=\frac{2 \mathrm{~L}}{\cos 	heta\left(\mathrm{~g} \sin 	heta-\mathrm{a}_0 \cos 	hetaight)}=\frac{2 \mathrm{~L}}{\mathrm{~g} \sin 	heta \cos 	heta-\mathrm{a}_0 \cos ^2 	heta} $$
Using the identity $2 \sin 	heta \cos 	heta=\sin 2 	heta$ and $\cos ^2 	heta=\frac{1+\cos 2 	heta}{2}$ :
$$ t^2=\frac{4 L}{2 g \sin 	heta \cos 	heta-2 a_0 \cos ^2 	heta}=\frac{4 L}{g \sin 2 	heta-2 a_0\left(\frac{1+\cos 	heta}{2}ight)} $$
$$\Rightarrow $$ $t^2=\frac{4 L}{g \sin 2 	heta-a_0(1+\cos 	heta)}$
$\Rightarrow \mathrm{t}=\sqrt{\frac{4 \mathrm{~L}}{\mathrm{~g} \sin 2 	heta-\mathrm{a}_0(1+\cos 	heta)}}$
Hence, the correct option is (B).

Answer

2

Answer

4

Answer

$\sqrt{2}$

Answer

$2 \sqrt{2}$

Laws of Motion

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