A stone of 1 kg is thrown with a velocity
Question.
A stone of 1 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 50 m. What is the force of friction between the stone and the ice ?

Solution:
Mass of stone, $\mathrm{m}=1 \mathrm{~kg}$; initial velocity of stone, $\mathrm{u}=20 \mathrm{~ms}^{-1}$; final velocity of stone, $\mathrm{v}=0 ;$ distance covered by the stone, $\mathrm{s}=50 \mathrm{~m} ;$ acceleration of stone, $\mathrm{a}=? ;$ force acting on the stone due to friction, $\mathrm{F}=$ ?

We know, $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}$

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

or $-400=100$ a

$a=\frac{-400}{100}=-4 \mathrm{~ms}^{-2}$

$\therefore$ Force of friction, $\mathrm{F}=\mathrm{ma}=1 \times(-4)=-4 \mathrm{~N}$

Negative sign signifies that force of friction is acting in the direction opposite to the direction of motion of the stone.
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