Question

A stone of 1 kg is thrown with a velocity of 20 m/s 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 ice.

Please reply fast

Answer 1

deceleration of stone =   a
v² - u² = 2 a s
0² - 20² =  2 a * 50  =>  a = - 4 m/s²
force = m  a  =  - 4 * 1  =  - 4 N

Answer 2

initial velocity (u)= 20 m/s
final velocity (v) = 0 m/s
distance covered (s) = 50 m
mass (m) = 1 kg
 Acc. to third eq. of motion
v2 = u2 + 2as

so on solving it
we get acceleration (a) = 4m/s2
so, force of friction (f) = 4 N