Answer
(a) $130J$
(b) $2200W$
(c) more than
Work Step by Step
(a) We can find the required work done as follows:
$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
We plug in the known values to obtain:
$W=\frac{1}{2}(0.14)(42.5)^-\frac{1}{2}m(0)^2$
$W=130J$
(b) We can find the required power as follows:
$P=\frac{W}{t}$
We plug in the known values to obtain:
$P=\frac{130}{0.060}$
$P=2200W$
(c) We know that the power produced in this case is more than that found in part (b) because the ball is accelerated to the same speed in less time. That is, the same amount of work is done but in a smaller amount of time.
