A constant force of friction of 50 N is acting on a body of mass 200kg moving initially with a speed of 15m/s. How long does the body take to stop? What distance will it cover before coming to rest?

Answer 1

M*g = 200*9.8 = 1960 N. = Wt. of the body. = Normal(Fn).

Fk = u*Fn = 50 N.
u * 1960 = 50.
u = 0.0255.

a = u*g = 0.0255 * (-9.8) = -0.25 m/s^2.

V = Vo + a*t = 0.
15 -0.25t = 0.

0.25t = 15.
t = 60 s.

d = Vo*t + 0.5a*t^2.
d = 15*t - 0.125*60^2 = 450 m.

Answer 2

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