Question

b) A stone of mass 3.5 kg is thrown with a velocity of 14 m/s across the frozen surface and comes to rest after travelling a distance of 70m. What is the force of friction between the stone and ice?

Answer 1

V = 0

U = 14m/s

T = Distance / Speed

= 70m/ 14ms¹

= 5 second

.

a=( v - u)/t

= (0 -14m/s)/5s

= -2.8m/s

F = M x a

= 3.5kg x 2.8m/s²

= 9.80kgm/s²

= 9.8N

Coefficient of friction = ma/mg = a/g = 2.8/10 = 0.28

Force of friction = Umg

0.28 X 35

= 9.8N approx..

Thanks...