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 speed of the block when it reaches the point

Q

(A) $5 m s^{- 1}$
(B) $10 m s^{- 1}$ ✓ Correct
(C) $103 m s^{- 1}$
(D) $20 m s^{- 1}$

Answer: (B)

Solution

From the figure, point Q is at 30° below the horizontal through the center. Height drop:

h = R sin 30° = 40 \times 0.5 = 20 m

.

By work-energy theorem:

W_{gravity} + W_{friction} = Δ K E

m g h - 150 = \frac{1}{2} m v^{2}

1 \times 10 \times 20 - 150 = \frac{1}{2} \times 1 \times v^{2}

200 - 150 = \frac{v ^{2}}{2}

v^{2} = 100

v = 10 m/s

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

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