Question 40: Laws of Motion

Question

A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is $\_\_\_\_$ N.
$$ \left(g=10 \mathrm{~m} / \mathrm{s}^2\right) . $$

Accepted Answer

The forces acting on the block are,
Component of weight down the plane (along x -axis) $=\mathrm{mg} \sin 	heta=5 	imes 10 	imes \sin 30^{\circ}=25 \mathrm{~N}$
Normal Force, $\mathrm{N}=\mathrm{mg} \cos 	heta$
Since the block moves down, friction acts up the plane.
$$ \mathrm{f}_{\mathrm{k}}=\mu_{\mathrm{k}} \mathrm{~N}=\mu(\mathrm{mg} \cos 	heta) $$
Maximum frictional force :
$$ \mathrm{f}_{\mathrm{k}}=\frac{\sqrt{3}}{2} 	imes\left(5 	imes 10 	imes \cos \left(30^{\circ}ight)ight) $$
$$\Rightarrow $$ $$ \mathrm{f}_{\mathrm{k}}=\frac{\sqrt{3}}{2} 	imes\left(50 	imes \frac{\sqrt{3}}{2}ight)=\frac{3}{4} 	imes 50=37.5 \mathrm{~N} $$
Since the upward frictional force ( 37.5 N ) is greater than the downward component of weight ( 25 N ), the block will not move down on its own. To make it move down at a constant velocity, an additional downward force (P) must be applied.
The equilibrium equation along the incline is :
Net Force Downward $=$ Net Force Upward
$$ \mathrm{P}+\mathrm{F}_{	ext {weight }}=\mathrm{f}_{\mathrm{k}} $$
$$\Rightarrow $$ $P+25=37.5$
$$ \mathrm{P}=37.5-25=12.5 \mathrm{~N} $$
Hence, the correct option is (B).

Answer

25

Answer

12.5

Answer

15

Answer

7.5

Laws of Motion

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