Question

a stone of 1kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after traveling a distance of 50 m what is the force of friction between the stone and the ice​

Answer 1

Answer:

4 N

Explanation:

u = 20m/s

s = 50 m

v = 0 m/s

m = 1 kg

We Know That

v^2 - u^2 = 2as

( 0 )^2 - ( 20 )^2 = 2 × a× 50

400 = 100 a

a = 4 ms-2

Therefore

F = ma

F = 1 × 4

F = 4 N