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 in figure. The velocities of the object at points x = 8 m and x = 12 m, would he the
respective values of (nearly) [NEET (Odisha) 2019]

The area under the 
force-displacement curve gives the amount of work done.
From work-energy theorem,  $W=\Delta \mathrm{KE}$     …(i)

$\therefore \text{ At } x=8 \mathrm{m}, W=\text{ Area }ABDO+\text{ Area }CEFD$

$=20\times 5+10\times 3=130 \mathrm{J}$

Given,    $m=500 \mathrm{g}=500\times {10}^{-3} \mathrm{kg}$

Using Eq. (i), we get

$\Rightarrow 130=\frac{1}{2}m{v}^2=\frac{1}{2}\times 500\times {10}^{-3}\times v^2$

$\Rightarrow v=2\sqrt{130}=22.8{ \mathrm{ms}}^{-1}\approx 23{ \mathrm{ms}}^{-1}$

$\text{ At } x=12 \mathrm{m}$

 $W=\text{ Area }ABDO+\text{ Area }CEFD$  $+\text{ Area }FGHIJ+\text{ Area }KLMJ$

$W=20\times 5+10\times 3+\left(-20\times 2+\frac{1}{2}\times (-5)\times 2\right)$  $+10\times 2$

$\because \text{ Area }FGHIJ=\text{ Area }FGIJ+\text{ Area }GHI)$

$=100+30-40-5+20=105 \mathrm{J}$

Using Eq. (i), we get

$\therefore 105=\frac{1}{2}\times 500\times {10}^{-3}\times v^2$

$\Rightarrow v=2\sqrt{105}\simeq 20.6{ \mathrm{ms}}^{-1}$