Question

A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of lake and comes to rest after traveling a distance bof 50m what is the force of friction b/w the stone and ice? ( follow me and I will follow you back promise to you friends )​

Answer 1

Answer:

4N

Explanation:

In this question we will use the equation of motion v^2 = u^2 +2as

where v=0,u=20,s=50,a=? To find a

As 0 =400+2a×50

400=-100a

4= -a

Now since f=ma

So f = 1×4 = 4N

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Answer 2

Before we proceed we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force is acting in leftward while the stone in moving rightward direction. One thing we have to fix in our mind that frictional force will always act opposite to the direction of motion of the body.

Now on resolving the forces,

f−ma or f=ma ------(1)

also form the equation of motion v² = u² +2as, here final velocity is zero,so v = 0

u² = −2as

a = -u²/2s = -20²/20×50 = -4m/sec²

on substituting the given data in equation(1)

f = 1 × (−4) = −4N

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