Question

5. A stone of 1 kg is thrown with a velocity of 20 m s' 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 1

Answer:

The initial velocity of the stone, u= 20 m/s

The 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 a we get,

F = 1 × (-4) = -4N

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