Question

An iron sphere of 1Kg is moving a velocity of 20m/s on a cemented floor. It comes to rest after travelling a distance Of 50m.Find the force of friction between the sphere and the floor(

Answer 1

Given:
Initial velocity (u) = 20m/s
final velocity= 0m/s
Distance=50 m
from, third equation of motion:
v² - u² = 2as
0² - 20² = 2 x a x50
a=-400/2x50
a=-4m/s²
From Newton's second law, F=ma
F=1x-4
   =-4N
Therefore, a frictional force of -4N ( negative sign indicates force acts in opposite direction.)

Answer 2

Answer:

Explanation:

Given:

Initial velocity (u) = 20m/s

final velocity= 0m/s

Distance=50 m

from, third equation of motion:

v² - u² = 2as

0² - 20² = 2 x a x50

a=-400/2x50

a=-4m/s²

From Newton's second law, F=ma

F=1x-4

  =-4N

Therefore, a frictional force of -4N ( negative sign indicates force acts in opposite direction.)