JEE Advanced 2013 · Paper 2 · Physics Q13

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PhysicsSingle correct+3 / -1Question 13 of 60

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Paragraph for Questions 13 and 14

A small block of mass 1 kg 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, $g=10\ \mathrm{m\,s^{-2}}$ ).

A quarter-circle track of radius R in the x-y plane, concave towards its centre P, which lies at height R directly above the track's lowest point (where it meets the x-axis); the track starts at the top on the y-axis, where the block is. Q is on the arc where the radius PQ makes 30° below the horizontal line through P.

The speed of the block when it reaches the point $Q$ is

5\ \mathrm{m\,s^{-1}}

10\ \mathrm{m\,s^{-1}}

10\sqrt3\ \mathrm{m\,s^{-1}}

20\ \mathrm{m\,s^{-1}}

Answer and worked solution

Official answer: B

Height lost. The block starts level with the centre P (height $R=40$ m). At Q the radius makes 30° below the horizontal, so Q is $R\sin30°=20$ m below the start.

Energy. $\frac12mv^2=mgh-W_f=1(10)(20)-150=50$ J, so $v=10$ m/s.

(B)

Topics:Work-energy theorem

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JEE Advanced 2013 · Paper 2, Physics question 13