An object of mass 500 \mathrm{~g}, initially at rest, is acted upon by a variable force whose X-component varies with X in the manner shown. The velocities of the object at the points \mathrm{X}=8 \mathrm{~m} and \mathrm{X}=12 \mathrm{~m}, would have the respective values of (nearly) (1) 18 \mathrm{~m} / \mathrm{s} and 20.6 \mathrm{~m} / \mathrm{s} (2) 18 \mathrm{~m} / \mathrm{s} and 24.4 \mathrm{~m} / \mathrm{s} (3) 23 \mathrm{~m} / \mathrm{s} and 24.4 \mathrm{~m} / \mathrm{s} (4) 23 \mathrm{~m} / \mathrm{s} and 20.6 \mathrm{~m} / \mathrm{s}
An object of mass 500 \mathrm{~g}, initially at rest, is acted upon by a variable force whose X-component varies with X in the manner shown. The velocities of the object at the points \mathrm{X}=8 \mathrm{~m} and \mathrm{X}=12 \mathrm{~m}, would have the respective values of (nearly) (1) 18 \mathrm{~m} / \mathrm{s} and 20.6 \mathrm{~m} / \mathrm{s} (2) 18 \mathrm{~m} / \mathrm{s} and 24.4 \mathrm{~m} / \mathrm{s} (3) 23 \mathrm{~m} / \mathrm{s} and 24.4 \mathrm{~m} / \mathrm{s} (4) 23 \mathrm{~m} / \mathrm{s} and 20.6 \mathrm{~m} / \mathrm{s}

Answer (4)
Sol.
Using work-energy theorem
\Delta K= work = area under F=x graph
From x=0 to x=8 m
\begin{array}{l} \frac{1}{2} m v^{2}=100+30 \\ v^{2}=520 \\ v=\sqrt{520}=23 \mathrm{~m} / \mathrm{s} \end{array}
From x=0 to x=12 \mathrm{~m}
\begin{array}{l} \frac{1}{2} m v^{2}=100+30-47.5+20 \\ v=\sqrt{410} \\ v=20.6 \mathrm{~m} / \mathrm{s} \end{array}
Hence appropriate option is 23 \mathrm{~m} / \mathrm{s} and 20.6 \mathrm{~m} / \mathrm{s}