Question

a stone of 3 kg is thrown with a velocity of 20 metre per second across the frozen surface of a lake and comes to the rest after travelling a distance of hundred metre what is the force of friction between the stone and ice​

Answer 1

Answer:

m = 1kg

u = 20m/s. v = 0m/s.

s(distance travelled) = 50m

using third equation of motion

v²=u²+2as

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

F = m×a

F = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction)

hope this helps