an iron sphere of 1kg is moving a velocoty of 20 m s-1on a cemented floor. It comes to rest after travelling a distance of 50m. Find the force of friction between the sphere and the floor.
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 (finally the stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v 2 = u 2 + 2 as
Where,
Acceleration, a
(0) 2 = (20) 2 + 2 × a × 50
a = −4 m/s 2
The negative sign indicates that acceleration is acting against the motion of the iron sphere.
Mass of the sphere, m = 1 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma
F = 1 × (− 4) = −4 N
Hence,the force of friction between the sphere and the floor is −4 N.