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PhysicsInteger+3Question 17 of 60

JEE Advanced 2014 · Paper 1, Physics question 17

Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n×10)(n\times10) Joules. The value of nn is (take acceleration due to gravity =10 m s−2=10\ \mathrm{m\,s^{-2}})

A quarter-elliptical rail in a vertical plane from P (3 m right of O, at ground level) up to Q (4 m above O); a block on the rail is pulled by a force parallel to the chord PQ (dashed), with the right angle at O.
Answer and worked solution

Official answer: 5

Work by the pulling force. The force is constant (18 N along PQ), so its work depends only on the displacement PQ→\overrightarrow{PQ}, which has length 32+42=5\sqrt{3^2+4^2}=5 m along the force:
WF=18×5=90 J.W_F=18\times5=90\text{ J}.

Work by gravity. The block rises 4 m: Wg=−1×10×4=−40W_g=-1\times10\times4=-40 J. The normal force does no work and there is no friction.

Kinetic energy at Q. Starting from rest, K=90−40=50K=90-40=50 J =5×10=5\times10 J.

Answer: 5

Topics:Work-energy theoremWork by constant and variable forces
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