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JIPMER 2012
Exam
JIPMER
Year
2012
Subject
Physics
Chapter
Laws of Motion

JIPMER 2012 · Physics · Laws of Motion

A stone weighing 1 kg and sliding on ice with a velocity of 2 ~m /...

A stone weighing 1 kg and sliding on ice with a velocity of 2 m/s2 \mathrm{~m} / \mathrm{s} is stopped by friction in 10 s . The force of friction (assuming it to be constant) will be

Options

  1. A-20 N
  2. B-0.2 N
  3. C0.2 N
  4. D20 N

Correct answer

B. -0.2 N

Step-by-step solution

u=2 m/s,v=0,f=0 su=2 \mathrm{~m} / \mathrm{s}, \quad v=0, \quad f=0 \mathrm{~s} a=v−uta=\frac{v-u}{t} =0−210=−210=\frac{0-2}{10}=-\frac{2}{10} =−15=-\frac{1}{5} =−0.2 m/s2=-0.2 \mathrm{~m} / \mathrm{s}^2 ∴\therefore Friction force =ma=m a =1×(−0.2)=−0.2 N=1 \times(-0.2)=-0.2 \mathrm{~N}
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