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CGP EDU Academic Team
Published on: September 12, 2026
A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s is stopped by friction in 10 sec. The force of friction (assuming it to be constant) will be
Text Solution
Verified by ExpertsThe correct answer is:
B
$u = 2 \text{ m/s}, v = 0, t = 10 \text{ sec}$
$\therefore a = \frac{v-u}{t} = \frac{0-2}{10} = -\frac{2}{10} = -\frac{1}{5} = -0.2 \text{ m/s}^2$
$\therefore$ Friction force $= ma = 1 \times (-0.2) = -0.2 \, \mathrm{N}$
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