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.

Asked by Vishal kumar | 2 years ago |  291

1 Answer

Solution :-

Given data:

Initial velocity u=5 ms–1

Mass of the body = 20kg

Final velocity v = 2 ms–1

The initial kinetic energy

Ei =\(( \frac{1}{2})\) mu2 = \(( \frac{1}{2})\) x 20 x (5 ms–1 ) 2 =250kgms-2

= 250Nm = 250J

Final kinetic energy Ef = \(( \frac{1}{2})\) mv2 = \(( \frac{1}{2})\) x 20 x (2 ms–1 ) 2 =40kgms-2 = 40 Nm =40J

Therefore,

Work done = Change in kinetic energy

Work done = Ef – Ei

Work done =40J – 250J

Work done = -210J

Where negative sign indicates that force acts contrary to motion direction.

Answered by Shivani Kumari | 2 years ago

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