Question

A stone of 1 kg is thrown with a velocity of 20 m s-1 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?

Answer 1

Given

$\bf Initial \: Velocity (u) = 20 \: m \: {s}^{ - 1}$

$\bf Final \: Velocity (v) = 0 \: m \: s^{ - 1}$

$\bf Distance \: Covered = 50 \: m$

Calculating Acceleration

As, we know

$\huge \boxed{ \bf \red{ {v}^{2} - {u}^{2} = 2as}}$

$\longrightarrow \bf( {0)}^{2} - ( {20) }^{2} = 2 \times a \times 50$

$\longrightarrow \bf0 - 400 = 100a$

$\longrightarrow \bf - 4 \cancel{00} = \cancel{100}a$

$\longrightarrow \bf a = - 4$

Thus, here the Acceleration is -4 m/s and the result is negative so, there is a Retardation

Therefore, we got Acceleration = -4m/s

As we know,

$\huge \boxed{ \bf \red{F = m \times a}}$

$\bf F = 1 \times - 4$

$\longrightarrow \bf- 4 \: N$

Thus, force of Friction between the stone and Ice is -4 N

Important Formula's

$\sf Force = Mass \times Acceleration$

$\sf v = u + at$

$\sf v^{2} - u^{2} = 2as$

$\sf S = ut + \dfrac{1}{2} at^2$

Answer 2

We have been told that :

$\longrightarrow{\rm{Initial \:Velocity (u) = {20ms}^{-1}}}$

$\longrightarrow{\rm{Final \: Velocity (v) = {0ms}^{-1}}}$

$\longrightarrow{\rm{Distance (s) = 50m}}$

Formula :

$\large{\boxed{\rm{{(v)}^{2} - {(u)}^{2} = 2as }}}$

Putting up the values :

$\longrightarrow{\rm{{(0)}^{2} - {(20)}^{2} = 2 \times a \times 50}}$

$\longrightarrow{\rm{0 - 400 = 100a}}$

$\longrightarrow{\rm{a = \dfrac{- 400}{100}}}$

$\longrightarrow{\rm{a = \cancel{ \dfrac{- 400}{100}}}}$

$\large{\boxed{\rm{ a = -4}}}$

So ,

Acceleration : -4m/s