JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Work, Power & Energy

Paragraph for Questions 13 and 14

A small block of mass

1 k g

is released from rest at the top of a rough track. The track is a 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}

).

The magnitude of the normal reaction that acts on the block at the point

Q

(A) $7.5 N$ ✓ Correct
(B) $8.6 N$
(C) $11.5 N$
(D) $22.5 N$

Answer: (A)

Solution

At point Q (30° below horizontal), the block has speed

v = 10

m/s.

For circular motion, the net radial force provides centripetal acceleration:

N - m g cos 60° = \frac{m v ^{2}}{R}

(The angle from the vertical at Q is 60°)

N = m g cos 60° + \frac{m v ^{2}}{R}

N = 1 \times 10 \times 0.5 + \frac{1 \times 100}{40}

N = 5 + 2.5 = 7.5 N

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Work, Power & Energy

Solution

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