Question

Stone size 5kg is thrown with a velocity of 16 m/s across the frozen of a lake and come to rest after traveling a distance 50 m . What is the force of friction between the stone and ice..​

Answer 1

Step-by-step 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

Answer:

Hii,

Step-by-step explanation:

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