Answer
(a) $43.5J$
(b) greater than
(c) $7.00\times 10^8J$
Work Step by Step
(a) We know that
$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$W=\frac{1}{2}m_{\circ}[v_f^2-v_i^2]$
We plug in the known values to obtain:
$W=\frac{1}{2}(0.145Kg)[(35.0m/s)^2-(25.0m/s)^2]=43.5J$
(b) We know that in the given scenario, the mass of the ball ($m_{\circ}$) remains the same. Thus, the velocities $v_f$ and $v_i$ are greater than in case (a) and therefore, the work done required is also greater than that determined in part(a).
(c) We know that
$W=m_{\circ}c^2[\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}]$
We plug in the known values to obtain:
$W=(0.145Kg)(3\times 10^8m/s)[\frac{1}{\sqrt{1-\frac{(200,000,035m/s)^2}{(3\times 10^8)^2}}}-\frac{1}{\sqrt{1-\frac{(200,000,025m/s)^2}{(3\times 10^8)^2}}}]$
This simplifies to:
$W=7.00\times 10^8J$
