Question

6. A stone of 1 kg is thrown with a velocity of 20 m s-l 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

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)