Physics - W.E.P. Question with Solution | TestHub

The area under the force displacement curve give the amount of work done.From work-energy theorem, $$\begin{array}{rl}& \mathrm{W}=\Delta \mathrm{KE}\\ & \therefore \text{ At }\mathrm{x}=8\mathrm{m}\text{, }\\ & \mathrm{W}=\text{ Area }\mathrm{ABDO}+\text{ Area CEFD }\\ & =20\times 5+10\times 3=130\mathrm{J}\\ & \text{ Using Eq. (i) }\\ & \Rightarrow 130=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\times \frac{500}{1000}{\mathrm{v}}^2\\ & \Rightarrow \mathrm{v}=2\sqrt{130}=22.8{\mathrm{ms}}^{-1}\approx 23{\mathrm{ms}}^{-1}\\ & \text{ At }\mathrm{x}=12\mathrm{m}\\ & \mathrm{W}=\text{ Area }\mathrm{ABDO}+\text{ Area CEFD + Area FGHIJ }\\ & +\text{ Area KLMJ }\\ & \mathrm{W}=20\times 5+10\times 3+(-20\times 2)+\left(\frac{1}{2}\times -5\times 2\right)\\ & +10\times 2[\because \text{ Area FGHIJ }=\text{ Area }\\ & \mathrm{FGIJ}+\text{ Area }\mathrm{GHI}]\\ & =100+30-40-5+20=105\mathrm{J}\\ & \text{ Using Eq. (i) }\\ & \therefore 105=\frac{1}{2}\times \frac{1}{2}\times v^2\\ & \Rightarrow \mathrm{v}=2\sqrt{105}\simeq 20.6{\mathrm{ms}}^{-1}\\ & \end{array}$$