JEE Main 2014 · 11 Apr · Shift 1 · Physics Q4

← JEE Main 2014 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 4 of 90

A small ball of mass m starts at a point A with speed $v_o$ and moves along a frictionless track AB as shown. The track BC has coefficient of friction $\mu$ . The ball comes to stop at C after travelling a distance L which is :

A curved frictionless track falls from A, at height h, to the horizontal ground at B; the rough horizontal stretch BC has length L.

\dfrac{2h}\mu+\dfrac{v_o^2}{2\mu g}

\dfrac h\mu+\dfrac{v_o^2}{2\mu g}

\dfrac h{2\mu}+\dfrac{v_o^2}{\mu g}

\dfrac h{2\mu}+\dfrac{v_o^2}{2\mu g}

Answer and worked solution

Official answer: B

On the rough stretch, friction does work $\mu mgL$ , which must remove all the kinetic energy at B:
$\mu mgL=mgh+\tfrac12mv_o^2\implies L=\frac h\mu+\frac{v_o^2}{2\mu g}.$

Topics:Work-energy theorem Friction

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 4