Chapter - 2 Force and Laws of Motion

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Q
Force and Laws of Motion

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?

Answer:

Given: m = 1 kg

u = 20 m/s

s = 50 m

Because the stone comes into rest after travelling 50 m distance, therefore the final velocity of the stone will be zero.

v = 0

To find: Force of friction (F) on the stone.

Using 3rd equation of motion

v^2 - u^2 = 2as

(0)^2 - (20)^2 = 2a(50)

-400 = 100 	imes a

a = - 4 m/s^2

Now, for the force (F)

F = ma

F = 1 kg 	imes - 4 m/s^2

F = -4 N

Therefore the force of friction between the stone and ice is -4 N.


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