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A stone of size $1 kg$ 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 $50 m$ . Find the force of friction between the stone and the ice.

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- 2 N

- 8 N

- 4 N

- 6 N

answer is C.

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Detailed Solution

Force of friction between the stone and the ice is

- 4 N

.
Given,
Mass of stone,

m = 1 kg

Its initial velocity,

u = 20 m / s

Final velocity,

v = 0 m / s

Distance travelled,

s = 50 m

From Newton’s third law of motion,

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

⇒

2 \times a \times 50

= {(0)}^{2} - {(20)}^{2}

⇒

100 a = - 400

⇒

a = - 4 m s^{- 2}

(negative sign shows retardation)
Force of friction between the stone and ice

=

Force required to stop the stone.

f = ma

= 1 \times (- 4)

= - 4 N

∴ Friction between the stone and the ice is

- 4 N

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