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


Answer :

Given,

Mass of the body = 20 kg

Initial velocity( u ) =5 m/s

Final velocity( v ) = 2 m/s

The initial kinetic energy

\(= E_i \ = \ \frac{mu^2}{2} \ \)
\( = \ \frac{20 x (5)^2}{2} \ \)
\( = \ 250 \ kgm/s^2 \)

\( \therefore E_i \ = \ 250 \ J \)

Final kinetic energy

\(= E_f \ = \ \frac{mv^2}{2} \ \)
\( = \ \frac{20 x (2)^2}{2} \ \)
\( = \ 40 \ kgm/s^2 \)

\(\therefore E_f \ = \ 40 \ J \)

\(\therefore Work \ done \ = \ Change \ in \ kinetic \ energy \)

\( \Rightarrow Work \ done \ = \ E_f \ – \ E_i \)

\( \Rightarrow Work \ done \ = \ 40 \ J \ – \ 250 \ J \)

\(\Rightarrow Work \ done \ = \ -210 \ J \)

Where negative sign indicates that force acts in opposite direction.

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