PERES LAWHow does Ampere's Law relate to electromagnetism?

AI Thread Summary
The discussion centers on calculating the heat input required for an engine operating at 26% efficiency, which raises a 6-kg crate to a height of 11 m while achieving a speed of 5 m/s. The initial calculation for work done was based solely on gravitational potential energy, yielding 647.46 J, but neglected the kinetic energy component. The correct approach involves combining both potential energy and kinetic energy to find the total work output. The participants emphasize the need to properly apply the efficiency formula to determine the heat input accurately. Clarifying these calculations will lead to the correct heat input value for the engine.
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Homework Statement



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?

Homework Equations



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

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
 
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mikefitz said:

Homework Statement



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?

Homework Equations



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

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
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
 
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?
 
mikefitz said:
for KE do I just say that delta E = 1/2mv^2 + mgh = W = eQ ?
Looks good to me.

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

mgh = 647.46J
What do i do with these two values?
Why not put them into your equation and find Q?

AM
 
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