JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Work Done by a Constant Force

Let a constant force \[\overrightarrow{F}\] be applied on the body such that it makes an angle q with the horizontal and body is displaced through a distance s.

By resolving force \[\overrightarrow{F}\] into two components :

(i) F cos \[\theta \] in the direction of displacement of the body.

(ii) F sin \[\theta \] in the perpendicular direction of displacement of the body.

Since body is being displaced in the direction of \[F\cos \theta \], therefore work done by the force in displacing the body through a distance s is given by

\[W=(F\cos \theta )\,s=Fs\cos \theta \] or                 \[W=\overrightarrow{F}.\overrightarrow{s\,}\]

Thus work done by a force is equal to the scalar (or dot product) of the force and the displacement of the body.

If a number of forces \[{{\overrightarrow{F}}_{1}},\,{{\overrightarrow{F}}_{2}},\,{{\overrightarrow{F}}_{3}}......{{\overrightarrow{F}}_{n}}\] are acting on a body and it shifts from position vector \[{{\overrightarrow{\,r}}_{1}}\] to position vector \[{{\overrightarrow{\,r\,}}_{2}}\] then \[W=({{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}+....{{\overrightarrow{F}}_{n}})\,.({{\overrightarrow{\,r}}_{2}}-{{\overrightarrow{\,r}}_{1}})\]