Question

A constant force of magnitude 50 N is applied to a body of 10 kg moving initially with a speed of 10 m s–1. How long will it take the body to stop if the force acts in a direction opposite to its motion?

Answer 1

I think this may help pls brainlist me if correct

Answer 2

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Retarding force,

F = –50 N

Mass of the body,

m = 10 kg

Initial velocity of the body,

u = 10 m/s

Final velocity of the body,

v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

F = ma

=> –50 = 10 × a

=> a= -50/10 = -5 m/s^2

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

=> t= -u/a = -10/-5= 2 sec.

I hope, this will help you

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