A block of mass M = 1 kg is released from re | vedclass

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Question

(A) $1.$ Let the block be released from height $h = R = 40 \ m$. The vertical height of point $Q$ from the bottom is $h_Q = R - R \sin(30^{\circ}) = R

Vedclass · Accepted Answer

$(B, A)$

Vedclass · Answer

(A) $1.$ Let the block be released from height $h = R = 40 \ m$. The vertical height of point $Q$ from the bottom is $h_Q = R - R \sin(30^{\circ}) = R - R/2 = R/2 = 20 \ m$.Applying the work-energy theorem between the starting point and point $Q$:$W_{gravity} + W_{friction} = \Delta K$$Mg(R - h_Q) - 150 = \frac{1}{2} M v^2$$1 	imes 10 	imes (40 - 20) - 150 = \frac{1}{2} 	imes 1 	imes v^2$$200 - 150 = 0.5 v^2 \Rightarrow 50 = 0.5 v^2 \Rightarrow v^2 = 100 \Rightarrow v = 10 \ m s^{-1}$.Thus,the speed at $Q$ is $10 \ m s^{-1}$. (Option $B$)$2.$ At point $Q$,the forces acting on the block are gravity $(Mg)$ and normal reaction $(N)$. The component of gravity perpendicular to the track is $Mg \cos(60^{\circ}) = Mg/2$.The net centripetal force is $N - Mg \cos(60^{\circ}) = \frac{M v^2}{R}$.$N - 1 	imes 10 	imes 0.5 = \frac{1 	imes 100}{40}$$N - 5 = 2.5 \Rightarrow N = 7.5 \ N$. (Option $A$)Therefore,the correct pair is $(B, A)$.

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