Question

A stone of size 5 kg. is thrown with a velocity of 40 m/sec.across the frozen Surface of a lake and comes to rest after travelling a distance of 50m. what is the force of friction between the stone and the ice?​

Answer 1

Answer:

-160 N

Explanation:

m = 5 kg

u = 40 m/s

v = 0 m/s

s = 50 m

t = s/u = 50/40 s

a = (v - u)/t = (0 - 40) (4/5) = (-8) (4) = -32 m/s²

f = ma = (5) (-32) = -160 N