Work Done by a Constant Force
Category : JEE Main & Advanced
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 \[W=({{\overrightarrow{F}}_{1}}+{{\overrightarrow{F}}_{2}}+{{\overrightarrow{F}}_{3}}+....{{\overrightarrow{F}}_{n}})\,.({{\overrightarrow{\,r}}_{2}}-{{\overrightarrow{\,r}}_{1}})\]
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