Question

Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1. Calculate the work done by the force.​

Answer 1

Given,

Mass, m = 20 kg

Initial velocity, u= 5 ms-1

Final velocity, v= 2 ms-1

Work done = change in K. E.

W = 1 half mv2 - 1 half mu2

= 1 half m ( v2 - u2)

= 1 half x 20 x (22 - 52)

= 10 (4-25)

= -10 x 21

= -210 J

Here, the negative sign implies the retarding nature of the applied force.

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Answer 2

_/\_Hello mate__here is your answer--

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Given:-

Mass of the body = 20 kg

Initial velocity = 5 m/s

Final velocity = 2 m/s

To find:- work done by body

Solution:-

We know that,

work done =Change in kinetic energy

= 1/2mu^2 - 1/2mv^2

= 1/2m(u^2 − v^2)

= 1/2(20)(5^2 − 2^2)

= 10(25 − 4)

= 210 J

I hope, this will help you.☺

Thank you______❤

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