Answer
(a) $F_T = 3010~N$
(b) $W_n = 7480~J$
(c) $W_T = 54,200~J$
(d) $W_G = -46,700~J$
(e) $v = 7.51 ~m/s$
Work Step by Step
(a) We can use a force equation to find the tension $F_T$ in the cable.
$\sum F = ma$
$F_T - mg = m(0.160g)$
$F_T = m(1.160g) = (265 ~kg)(1.160)(9.80 ~m/s^2)$
$F_T = 3010~N$
(b) The net work $W_n$ is $(\sum F)(d)$.
$W_n = (\sum F)(d)$
$W_n = mad$
$W_n = (265 ~kg)(0.160)(9.80 ~m/s^2)(18.0 ~m)$
$W_n = 7480~J$
(c) $W_T = (3010 ~N)(18.0 ~m)$
$W_T = 54,200~J$
(d) $W_G = - (265 ~kg)(9.80 ~m/s^2)(18.0 ~m)$
$W_G = -46,700~J$
(e) We can use kinematics or work-energy ideas to find the final speed. Let's use work-energy ideas to find the final speed:
$\Delta KE = W_n$
$\frac{1}{2}mv^2 = 7480 ~J$
$v = \sqrt{\frac{(2)(7480 ~J)}{265 ~kg}} = 7.51 ~m/s$
