A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. The Force of friction between the stone and the ice is ____ N.

A stone of

1 kg

is thrown with a velocity of

20 m / s

across the frozen surface of a lake and comes to rest after travelling a distance of

50 m

. The Force of friction between the stone and the ice is

- 4 N

.
Given:
Mass,

m = 1 kg

Initial velocity,

u = 20 m / s

The final velocity,

v = 0 m / s

Distance travelled,

s = 50 m

From the third equation of motion,

v^{2} = u^{2} + 2 as

\Rightarrow a = \frac{v^{2} - u^{2}}{2 s}

\Rightarrow

= \frac{0 - {(20)}^{2}}{2 \times 50}

\Rightarrow

= \frac{- 400}{100}

a = - 4 m / s^{2}

We know,
Force of friction between the stone and the ice is

= mass \times acceleration

= 1 kg \times (- 4) m / s

= - 4 N

Negative sign indicates that the direction of the force of friction is opposite to the direction of moving stone.

A stone of $1 kg$ is thrown with a velocity of $20 m / s$ across the frozen surface of a lake and comes to rest after travelling a distance of $50 m$ . The Force of friction between the stone and the ice is ____ $N$ .