Q120 / 160 · JIPMER 2012 · Physics

A stone weighing 1 kg and sliding on ice with a velocity of $2 \mathrm{~m} / \mathrm{s}$ is stopped by friction in 10 s . The force of friction (assuming it to be constant) will be

  • A
    -20 N
  • Correct answer
    B
    -0.2 N
  • C
    0.2 N
  • D
    20 N

Solution

$u=2 \mathrm{~m} / \mathrm{s}, \quad v=0, \quad f=0 \mathrm{~s}$ $a=\frac{v-u}{t}$ $=\frac{0-2}{10}=-\frac{2}{10}$ $=-\frac{1}{5}$ $=-0.2 \mathrm{~m} / \mathrm{s}^2$ $\therefore$ Friction force $=m a$ $=1 \times(-0.2)=-0.2 \mathrm{~N}$

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

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