# Work of a heat engine

1. Dec 6, 2006

### mikefitz

1. The problem statement, all variables and given/known data

An engine works at 26% efficiency. The engine raises a 6-kg crate from rest to a vertical height of 11 m, at which point the crate has a speed of 5 m/s. How much heat input is required for this engine?

2. Relevant equations

e=work output / heat input = W/Qin
W=Fd

3. The attempt at a solution

I assume the velocity is insignificant since W=Fd; W=(6*.981)*(11)= 647.46J

.26=647.46J/Qin
Qin=.0004016 J

the value for calculated heat is wayy too low; where did I screw up? Thanks

2. Dec 7, 2006

### Andrew Mason

For starters, you have calculated 1/Qin not Qin.

You are also ignoring the work that is required to give the load the kinetic energy. Add that to your calculation for W and then do the algebra properly and you will be fine.

AM

3. Dec 7, 2006

### mikefitz

for KE do I just say that delta E = 1/2mv^2 + mgh = W = eQ ?

.5(6kg)(5^2)= 75J

mgh = 647.46J
What do i do with these two values?

4. Dec 7, 2006

### Andrew Mason

Looks good to me.

Why not put them into your equation and find Q?

AM