Question

A stone of mass 2 kg is thrown with a velocity 5 m/s across the frozen surface of a lake and comes

to rest after travelling a distance of 25 m. what is the force of friction between the stone and ice?​

Answer 1

Answer:

FORCE OF FRICTION = -4N

Explanation:

Initial velocity of the stone, u= 20 m/s

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

We know that

F = m×a

Substituting above obtained value of a=-4 in F= m x awe get,

F = 1×(-4) = -4N

(Here the negative sign indicates the opposing force which is Friction)

Answer 2

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