JEE Main Physics question
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
Answer: (B)
Solution
Deceleration a = (0−2)/10 = −0.2 m/s². Force = ma = 1×(−0.2) = −0.2 N (negative indicating it opposes the direction of motion).