=
50MW is our net output.
Then, we plug it in.
= = 0.333 or 33.33%
W net, out = Qh - Ql
Ql = Qh - W net,out
Plug the values in.
Then, it becomes:
= 150MW - 50 MW
= 100MW
Thus, the cycle thermal efficiency is 33.33% and the heat rejected is 100MW.
Consult the attached free body diagram.
By Newton's second law, the net force on the crate acting parallel to the surface is
∑ F[para] = (370 N) cos(-20°) - f = 0
(this is the x-component of the resultant force)
where
• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force
• f = magnitude of kinetic friction
The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.
Solve for f :
f = (370 N) cos(-20°) ≈ 347.686 N
The net force acting perpendicular to the surface is
∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0
(this is the y-component of the resultant force)
where
• n = magnitude of normal force
• 1480 N = weight of the crate
• (370 N) sin(-20°) = magnitude of the vertical component of push
The crate doesn't move up or down, so it's also in equilibrium in this direction.
Solve for n :
n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N
Then the coefficient of kinetic friction is µ such that
f = µn ⇒ µ = f/n ≈ 0.216
