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

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Answer:

F=-50N

M=200kg

U=15m/s

F=Ma

a=F/M=-50/200=-0.25

V^2-U^2=2aS

0-(15)^2=2(-0.25)S

S=-225/-0.5=450m

V=U+at

0=15-0.25t

t=-15/-0.25=60s

Step-by-step explanation:

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Have a great day.

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User Fredmaggiowski
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