Q153 / 170 · Manipal (MET) 2016 · Physics

A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m . Speed of the block at the bottom is $2 \mathrm{~m} / \mathrm{s}$. Work done by the frictional force on the block when it reaches at the bottom is Figure

  • A
    8 J
  • Correct answer
    B
    -8 J
  • C
    4 J
  • D
    0 J

Solution

Given, $m=1 \mathrm{~kg}$, $\begin{aligned} v & =2 \mathrm{~m} / \mathrm{s} \\ R & =1 \mathrm{~m} \\ W & =?\end{aligned}$ From energy conservation, $m g h+W=\frac{1}{2} m v^2$ $\begin{aligned} W & =\frac{1}{2} m v^2-m g h \\ & =\frac{1}{2} \times 1(2)^2-1 \times 10=-8 \mathrm{~J}\end{aligned}$

Hence, the correct answer is option: 'B'.

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