JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Work Done Calculation by Force Displacement Graph

Let a body, whose initial position is \[{{x}_{i}}\], is acted upon by a variable force (whose magnitude is changing continuously) and consequently the body acquires its final position \[{{x}_{f}}\].  

Let F be the average value of variable force within the interval dx from position x to (x + dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in displacing it from position \[{{x}_{i}}\] to \[{{x}_{f}}\] will be equal to the sum of areas of all the such strips

\[dW=\overrightarrow{F}\,dx\]

\[\therefore \,W=\int_{{{x}_{i}}}^{{{x}_{f}}}{dW=\int_{{{x}_{i}}}^{{{x}_{f}}}{F\,dx}}\]

\[\therefore \,W=\int_{{{x}_{i}}}^{{{x}_{f}}}{(\text{Area}\,\text{of}\,\text{strip}\,\text{of}\,\text{width}\,dx)}\]

\[\therefore \,W=\text{Area}\,\text{under}\,\text{curve}\,\text{between}\,{{x}_{i}}\,\text{and}\,{{x}_{f}}\]

i.e. Area under force-displacement curve with proper algebraic sign represents work done by the force.