Question 345: Work, Energy and Power & Circular Motion

Question

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/s
2
)

Accepted Answer

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
$${N_Q} - mg\sin 30^\circ =  {{m{v^2}} \over R}$$
$$ \Rightarrow $$ $${N_Q}  =  mg\sin 30^\circ +{{m{v^2}} \over R}$$
$$ = 1 	imes 10 	imes {1 \over 2} + {{100} \over {40}}$$
$$ = 5 + {5 \over 2} = {{15} \over 2}$$
$$ \Rightarrow {N_Q} = 7.5$$ N
Hence, option (A) is correct.

Answer

7.5 N

Answer

8.6 N

Answer

11.5 N

Answer

22.5 N

Work, Energy and Power & Circular Motion

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