Efficiency of an Engine and Thermodynamics. Help me with this Problem!?

Question

An engine works at 28.0% efficiency. The engine raises a 5.00 kg crate from rest to a vertical height of 10.0 m, at which point the crate has a speed of 5.00 m/s. How much heat input is required for this engine?

______Joules

Answer 1

If an engine is at 28% efficiency, then 0.28 of heat input gets converted into work. Now, the engine does a specific amount of work, which we will calculate here:

mass: 5.0 kg
displacement: 10.0 m
speed: 5 m/s
W = Force x Distance
F = Mass x Acceleration
(Vf - Vi)^2 = Acceleration x (Displacement)^2
1.) By using the third equation, we get:
(5)^2 = a x (10)^2
So, acceleration is 0.4 m/s^2

2) Calculate the amount of force taken to move the crate by using the second equation:
F = m x a
F = 5.0 x 0.4
The force is 2.0 Newtons

3.) Use the first equation to calculate the total work for the engine:
W = f x d
W = 2.0 x 10.0
W = 20.0 Joules

Finally we have to compute for the amount of heat input since the engine is only working at 28% efficiency. This simply means that 0.28 times heat input equals to work. Let's do this:

(0.28) x Q = W
(0.28) x Q = 20.0
Q= 71.43 Joules

So, the amount of heat input is 71.43 Joules. Thanks.

Answer 2

the crate starts with no energy, and at the end point the crate has two kinds of energy
potential energy = mass x gravity x height
Kinetic energy = 1/2 mass x velocity squared
once you know the energy output by the engine (= the energy added to the crate), the energy put into the engine is, input = output/.28