Answer
(a) $8W_{\circ}$
(b) II.
Work Step by Step
(a) We know that
$W_{\circ}=\Delta K.E$
$\implies W_{\circ}=\frac{1}{2}m(v_f^2-v_i^2)$
$\implies W_{\circ}=\frac{1}{2}m[(50)^2-(0)^2]=\frac{1}{2}m(50)^2$
Now we can find the required work done as
$W=\frac{1}{2}m[(150)^2-(50)^2]=\frac{1}{2}m(50)^2[(3)^2-(1)^2]$
$W=W_{\circ}(9-1)=8W_{\circ}$
(b) We can see that the final speed $(150\frac{Km}{h})$ is three times the speed produced by the work $W_{\circ}$. Hence, the best explanation is option II.
