Question

A constant force of magnitude 50 n is applied on a body of mass 10 kg moving initially with a speed of 10 ms-1 against the direction of motion. how long will it take the body to come to rest?

Answer 1

F= ma
50=10xa

a=5m/s2

the deaccleration is 5.0m/s2

find out time to stop
v=u+at

0=10-5t
t=10/5
t=2sec

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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