An iron sphere of one kg is moving a velocity of 20m/son a cemented floor. It comes to rest after traveling a distance of 50m.Find the force of friction between the sphere and the floor.

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Solution

v^2 = u^2+2as
initial velocity u= 20m/s
distance s= 50m
final velocity v= 0m/s
a = -400/100 = -4m/s^2
force = ma = 1 -4
= -4 Newton

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An iron sphere of one kg is moving a velocity of 20m/son a cemented floor. It comes to rest after traveling a distance of 50m.Find the force of friction between the sphere and the floor.

v^2 = u^2+2as initial velocity u= 20m/s distance s= 50m final velocity v= 0m/s a = -400/100 = -4m/s^2 force = m*a = 1* -4 = -4 Newton

v^2 = u^2+2as
initial velocity u= 20m/s
distance s= 50m
final velocity v= 0m/s
a = -400/100 = -4m/s^2
force = ma = 1 -4
= -4 Newton