Question

A constant force of magnitude 50 N is applied on a body of mass 10 kg moving initially with a speed of 10 m s1 against the

Answer 1

The question is incomplete.
m = 10 kg
F = m a =50 N
          a = 50 /10 = 5 m/sec^2

u = 10 m/sec

The instantaneous velocity of the body = v = 10 + 5 t

distance traveled = s = 10 t + 1/2 5 t^2
Work done by the force :  1/2 m v^2 - 1/2 m * 10^2
                                = 5 (v^2 - 100)  J
                = 5(100 + 25 t^2 + 30 t - 100)  J
                 = 25 (5 t + 6) t   Joules
Power = 250 t + 150  Watts

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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