An object of mass 500 ~g, initially at rest acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the…

The area under the force displacement curve give the amount of work done. From work-energy theorem, $$\begin{aligned} & \mathrm{W}=\Delta \mathrm{KE} \\ & \therefore \text { At } \mathrm{x}=8 \mathrm{~m} \text {, } \\ & \mathrm{W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD } \\ & =20 \times 5+10 \times 3=130 \mathrm{~J} \\ & \text { Using Eq. (i) } \\ & \Rightarrow \quad 130=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \times \frac{500}{1000} \mathrm{v}^2 \\ & \Rightarrow \quad \mathrm{v}=2 \sqrt{130}=22.8 \mathrm{~ms}^{-1} \approx 23 \mathrm{~ms}^{-1} \\ & \text { At } \quad \mathrm{x}=12 \mathrm{~m} \\ & \mathrm{~W}=\text { Area } \mathrm{ABDO}+\text { Area CEFD + Area FGHIJ } \\ & + \text { Area KLMJ } \\ & \mathrm{W}=20 \times 5+10 \times 3+(-20 \times 2)+\left(\frac{1}{2} \times-5 \times 2\right) \\ & +10 \times 2 \quad[\because \text { Area FGHIJ }=\text { Area } \\ & \mathrm{FGIJ}+\text { Area } \mathrm{GHI}] \\ & =100+30-40-5+20=105 \mathrm{~J} \\ & \text { Using Eq. (i) } \\ & \therefore \quad 105=\frac{1}{2} \times \frac{1}{2} \times v^2 \\ & \Rightarrow \quad \mathrm{v}=2 \sqrt{105} \simeq 20.6 \mathrm{~ms}^{-1} \\ & \end{aligned}$$​W=ΔKE∴ At x=8 m, W= Area ABDO+ Area CEFD =20×5+10×3=130 J Using Eq. (i) ⇒130=21​mv2=21​×1000500​v2⇒v=2130​=22.8 ms−1≈23 ms−1 At x=12 m W= Area ABDO+ Area CEFD + Area FGHIJ + Area KLMJ W=20×5+10×3+(−20×2)+(21​×−5×2)+10×2[∵ Area FGHIJ = Area FGIJ+ Area GHI]=100+30−40−5+20=105 J Using Eq. (i) ∴105=21​×21​×v2⇒v=2105​≃20.6 ms−1​