#### Answer

(a) $W_g = -25,700~J$
(b) $W_T = 25,900~J$

#### Work Step by Step

(a) We can find the work done by gravity.
$W_g = F\cdot d$
$W_g = F ~d~cos(\theta)$
$W_g = (mg) ~d~cos(180^{\circ})$
$W_g = (750~kg)(9.80~m/s^2)(3.5~m)(-1)$
$W_g = -25,700~J$
(b) We can use the work energy theorem to find the work done by tension.
$KE_1+W_{ext} = KE_2$
$KE_1+W_T+W_g = KE_2$
$W_T = KE_2 -KE_1 - W_g$
$W_T = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2- W_g$
$W_T = \frac{1}{2}m(v_2^2-v_1^2)- W_g$
$W_T = \frac{1}{2}(750~kg)[(0.75~m/s)^2-(0.25~m/s)^2]- (-25,700)$
$W_T = 25,900~J$