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 and how far does the body take to stop?

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# 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 and how far does the body take to stop?

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## Dhiraj

F = -50 N

m = 200 kg

u = 15 m/s

v = 0 m/s

a =

^{F}/_{m}=^{-50}/_{200}=-0.25 m/s²Using,

2aS = v²-u²

⇒2×(-0.25)S = 0²-15²

⇒-0.50 S = -225

⇒S =

450 mNow,

v = u + at

⇒0 = 15 + (-0.25)t

⇒0.25t = 15

⇒t =

60 s=1 minSo, the body will stop at 450 m after 60 sec.