An object of mass $500 \mathrm{~g}$, initially at rest acted upon by a variable force whose $\mathrm{X}$ component varies with $\mathrm{X}$ in the manner shown. The velocities of the object a point $X=8 \mathrm{~m}$ and $X=12 \mathrm{~m}$, would be the respective values of (nearly)

• A

  $18 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$
• B

  $23 \mathrm{~m} / \mathrm{s}$ and $24.4 \mathrm{~m} / \mathrm{s}$
• Correct answer
  C

  $23 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$
• D

  $18 \mathrm{~m} / \mathrm{s}$ and $20.6 \mathrm{~m} / \mathrm{s}$

Solution

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

Hence, the correct answer is option: 'C'.