Velocity-time graph for a body of mass $10 kg$ is shown in figure. Work-done on the body in first two seconds of the motion is

 

Here, $\mathrm{m}=10 \mathrm{kg}, \mathrm{t}=2 \mathrm{s}$

$\mathrm{u}=50 \mathrm{m} {\mathrm{s}}^{-1} \mathrm{at} \mathrm{t}=0 \mathrm{s}$

$\mathrm{a}=\frac{∆\mathrm{\nu }}{∆\mathrm{t}}=\frac{50-0}{0-10}=-5 {\mathrm{ms}}^{-2}$

Speed of the body at $\mathrm{t}=2 \mathrm{s}$

${\text{ν=u+at=50+(-5)×2=40 ms}}^{-1}$

Using work energy theorem, 

Work done on the body = change in kinetic energy of the body

$∆\mathrm{W}=\frac{1}{2}{\mathrm{m\nu }}^2-\frac{1}{2}{\mathrm{mu}}^2=\frac{1}{2}\times 10\times {40}^2-\frac{1}{2}\times 10\times {50}^2$

       $=5\times (40-50)(40+50)=-4500 \mathrm{J}$