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JEE Advance - Physics (2013 - Paper 2 Offline - No. 5)

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 speed of the block when it reaches the point Q is
5 ms-1
10 ms-1
$$10\sqrt 3 $$ ms-1
20 ms-1

Explanation

JEE Advanced 2013 Paper 2 Offline Physics - Work Power & Energy Question 20 English Explanation

Here we will use the concept of work-energy theorem. By work-energy theorem,

Total work done = Change in KE

$$\Rightarrow$$ Workdone by gravity + workdone by friction = $$\Delta KE$$

$$ \Rightarrow mgR\sin 30^\circ - 150 = {1 \over 2}m{v^2} - 0$$

$$ \Rightarrow 1 \times 10 \times 40 \times {1 \over 2} - 150 = {1 \over 2}{v^2}$$

$$ \Rightarrow 200 - 150 = {1 \over 2}{v^2}$$

$$ \Rightarrow {v^2} = 100$$

$$ \Rightarrow v = 10$$ m/s

Hence, option (B) is correct.

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