Question

a stone of 200 gram is thrown with a velocity of 100 second across the frozen surface of a lake and cross to rest after travelling a distance of 10 m/s
what is the force of friction between stone and ice​

Answer 1

Answer:

— What is the force of friction between the stone and the ice? Initial velocity of the stone, u= 20 m/s. Final velocity of the stone, v= 0. Distance covered

Answer 2

Answer:

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)