Question

A constant retarding force of 50N is applied to a body of 20 kg moving initially with a speed of 15m/s. How long does it take to stop

Answer 1

Initial velocity ( u) = 15 m/s.
Final velocity ( v) = 0 m/s .

Force ( F) = - 50 N

Mass ( m ) = 20 kg.

We know that, By Newton's Second Laws of Motion :

F = m * a.

-50 = 20 * a

a =- 50/20 = - 2.5 m/s²

We know that a = v - u / t

-2.5 = 0 - 15 / t

-2.5 = -15 / t

t = 15 / 2.5 = 6

It takes 6 seconds to stop

Answer 2

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Retarding force,

F = –50 N

Mass of the body,

m = 20 kg

Initial velocity of the body,

u = 15 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 = 20 × a

=> a= -50/20 = -2.5 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 = -15/-2.5= 6 sec.

I hope, this will help you

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