A stone of 1 kg is thrown with a velocity of 20 ms−1 on

Question

A stone of $1 k g$ is thrown with a velocity of $20 m s^{- 1}$ on the frozen surface of a lake. The stone comes to rest after travelling a distance of $50 m$ . What is the force of friction between the stone and the ice?

Options:

A .

5 N

B .

3 N

C . There is no friction between stone and ice

D .

4 N

Answer: Option D
:
D

Initial velocity of the stone = $20 m s^{- 1}$ .
Final velocity of the stone, $v = 0$ (finally the stone comes to rest).
Distance covered by the stone = 50 m.
According to third equation of motion:
$v^{2} = u^{2} + 2 a s$
0 × 0 = 20 × 20 + 2 × a × 50
$\Rightarrow$ $a = - 4 m s^{- 2}$ .
The negative sign indicates that acceleration is in the opposite direction to the motion of the stone.
Mass of the stone, $m = 1 k g$ .
From the newton's second law of motion:
Force = mass × acceleration. Therefore,
$F = 1 \times - 4 = - 4 N$ .
Hence, the force of friction between the stone and the ice is 4N. The negative sign indicates that friction oppposes the relative motion.