Question

A constant retarding force of 60N is applied to a body of mass 20kg moving initially with a speed of 30m/s . How long does the body take to stop?

Answer 1

F = mass x acceleration Since, F = 60 N and. m = 20 kg Therefore, 60 = 20 x a a = 3m/s2 Given, retardation force ,therefore a = -3m/s2 v= 0 m/s (since body comes to rest) u = 30 m/s We know that a = (v-u)/t -3 = ( 0 - 30 ) / t ∴ -3t = -30 t = -30/-3 ∴ t = 10s Ans: The time taken by the body to stop is 10s.

Answer 2

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Retarding force,

F = –60 N

Mass of the body,

m = 20 kg

Initial velocity of the body,

u = 30 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

=> –60 = 20 × a

=> a= -60/20 = -3 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 = -30/-3= 10 sec.

I hope, this will help you

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