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JEE Advance - Physics (2013 - Paper 2 Offline - No. 4)

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s2)

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s2)

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s2)

The magnitude of the normal reaction that acts on the block at the point Q is
7.5 N
8.6 N
11.5 N
22.5 N

Explanation

JEE Advanced 2013 Paper 2 Offline Physics - Work Power & Energy Question 19 English Explanation

At point Q, the forces acting on the particle are normal reaction $N_Q$, gravitational force $m g$ and frictional force $f$. Resolve these in the directions parallel and perpendicular to the path. The net force towards the centre $(P)$ provides the centripetal acceleration. Apply Newton's second law in a direction normal to the path to get

$${N_Q} - mg\sin 30^\circ = {{m{v^2}} \over R}$$

$$ \Rightarrow $$ $${N_Q} = mg\sin 30^\circ +{{m{v^2}} \over R}$$

$$ = 1 \times 10 \times {1 \over 2} + {{100} \over {40}}$$

$$ = 5 + {5 \over 2} = {{15} \over 2}$$

$$ \Rightarrow {N_Q} = 7.5$$ N

Hence, option (A) is correct.

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