Answer
The change in mechanical energy is $~~8600~J$
Work Step by Step
We can find the distance the block moves:
$v_f^2 = v_0^2+2ad$
$d = \frac{v_f^2 - v_0^2}{2a}$
$d = \frac{(30~m/s)^2 - (10~m/s)^2}{(2)(2.0~m/s^2)}$
$d = 200~m$
We can find the change in kinetic energy:
$\Delta K = \frac{1}{2}m~(v_f^2-v_0^2)$
$\Delta K = \frac{1}{2}(15~kg)~[(30~m/s)^2-(10~m/s^2)^2]$
$\Delta K = 6000~J$
We can find the change in gravitational potential energy:
$\Delta U = U_f-U_0$
$\Delta U = mgh-0$
$\Delta U = mgd~sin~\theta$
$\Delta U = (15~kg)(9.8~m/s^2)(200~m)~(sin~5.0^{\circ})$
$\Delta U = 2600~J$
We can find the total change in mechanical energy:
$\Delta E = 6000~J+2600~J = 8600~J$
The total change in mechanical energy is $~~8600~J$
