A block of mass 2 kg is placed on inclined rough surface AC (as shown in figure) of coefficient of friction μ. If g = 10 m s ⁻², the net force (in N) on the block will be: (a) 10 √(3) (b) zero (c) 10 (d) 20

Answer: (b). Angle of repose: tan θr = μ = 1/√(3) ⇒ θr = 30°. The incline is at 30°, so mgsin 30° = 10 N equals μ mgcos 30° = 10 N. Friction balances the pull down the slope; the block stays at rest. So the answer is (b) zero.

A block of mass 2 kg is placed on inclined rough surface AC (as shown… | NEET 2023

Q12NEET 2023Easy

Friction on an Incline

A block of mass 2 kg is placed on inclined rough surface AC (as shown in figure) of coefficient of friction $\mu$ . If g = 10 m s ${}^{-2}$ , the net force (in N) on the block will be:

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Answer: (b)

Angle of repose: $\tan\theta_r = \mu = \dfrac{1}{\sqrt{3}} \Rightarrow \theta_r = 30^\circ$ .
The incline is at $30^\circ$ , so $mg\sin 30^\circ = 10\text{ N}$ equals $\mu mg\cos 30^\circ = 10\text{ N}$ .
Friction balances the pull down the slope; the block stays at rest.
So the answer is (b) zero.

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