Question

A stone of 1 kg is thrown with a velocity of $20\frac{m}{s}$ across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

• A. 5N
• B. 3N
• C. There is no friction between stone and ice
• D. 4N

Answer 1

The correct option is D

4N

Initial velocity of the stone = $20\frac{m}{s}$.

The 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:

$v^2=u^2+2as$

$0\times 0=20\times 20+2\times a\times 50$

$a=-4\frac{m}{s^2}$.

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m = 1 kg.

From Newton's second law of motion:

Force = mass × acceleration. Therefore,

$F=1\times (-4)=-4N$

Hence, the force of friction between the stone and the ice is 4 N. The negative sign above indicates that the direction of friction is opposite to the direction of motion.