Answer
The power due to the force applied to the block by the cable is $~~17,000~W$
Work Step by Step
We can find the angle of the incline above the horizontal:
$tan~\theta = \frac{d_2}{d_1}$
$\theta = tan^{-1}~(\frac{d_2}{d_1})$
$\theta = tan^{-1}~(\frac{30~m}{40~m})$
$\theta = 36.87^{\circ}$
Since the speed of the block is constant, the force on the block due to the cable is equal in magnitude to the sum of the frictional force and the component of the block's weight directed down the incline.
We can find the force on the block due to the cable:
$F = mg~sin~\theta+mg~cos~\theta~\mu_k$
$F = mg~(sin~\theta+\mu_k~cos~\theta)$
$F = (1400~kg)(9.8~m/s^2)~(sin~36.87^{\circ}+0.40~cos~36.87^{\circ})$
$F = 12,622~N$
We can find the power due to the force applied to the block by the cable:
$P = F~v$
$P = (12,622~N)(1.34~m/s)$
$P = 17,000~W$
The power due to the force applied to the block by the cable is $~~17,000~W$
