Question 69: Laws of Motion

Question

In the given arrangement of a doubly inclined plane two blocks of masses $$M$$ and $$m$$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is 0.25. The value of $$m$$, for which $$M=10 \mathrm{~kg}$$ will move down with an acceleration of $$2 \mathrm{~m} / \mathrm{s}^2$$, is: (take $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$ and $$\left.	an 37^{\circ}=3 / 4ight)$$

Accepted Answer

For M block
$$\begin{aligned}
& 10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 	imes 2 \
& \mathrm{~T}=80-15-20 \
& \mathrm{~T}=45 \mathrm{~N}
\end{aligned}$$
For $$\mathrm{m}$$ block
$$\begin{aligned}
& \mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} 	imes 2 \
& 45=10 \mathrm{~m} \
& \mathrm{~m}=4.5 \mathrm{~kg}
\end{aligned}$$

Answer

4.5 kg

Answer

6.5 kg

Answer

9 kg

Answer

2.25 kg

Laws of Motion

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