JIPMER2012PhysicsLaws of Motion
A stone weighing 1 kg and sliding on ice with a velocity of 2 ~m / s is stopped by friction in 10 s . The force of friction (assuming it to be constant) will be
Options
- A-20 N
- B-0.2 N
- C0.2 N
- D20 N
Correct answer
B. -0.2 N
Step-by-step solution
u=2 ~m / s , v=0, f=0 ~s a= v-u t = 0-2 10 =- 2 10 =- 1 5 =-0.2 ~m / s ^2 Friction force =m a=1 (-0.2)=-0.2 ~N