Question

A stone of 2 kg is thrown with a velocity of 20 ${ms}^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of 100 m. Calculate the frictional force between the stone and the ice.

• A. 2N
• B. (-) 2 N
• C. 3N
• D. 4N

Answer 1

The correct option is D

4N

Mass of stone (m) = 2 kg

Initial velocity of stone (u) = 20 ${ms}^{-1}$

Final velocity of stone (v) = 0

Distance covered by the stone (s) = 100 m

Acceleration of stone (a) =?

Force acting on the stone due to friction (F) = ?

We know: $v^2$ – $u^2$ = 2as

$0^2$ – ${20}^2$ = 2a $\times$ 100

– 400 = 200 a

a = -2 ${ms}^{-2}$

Force = m $\times$ a = 2 $\times$(-2) = -4N
The negative sign shows that the direction of friction is opposite to direction of motion.