Question

A stone of 1 kg is thrown with a velocity of20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. [a] what is the acceleration on the stone [b] what is the force of friction between the stone and ice? [c] how long it takes to come to rest? [d] If the same stone is thrown with a velocity of 40 m/s what will be the velocity at 50th meter? ,pls urgent for exams pls​

Answer 1

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 a we get,

F = 1 × (-4) = -4N

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

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