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Homework Help Overview

The problem involves calculating the heat input required for an engine operating at 26% efficiency that raises a 6-kg crate to a height of 11 m while achieving a speed of 5 m/s. The context is rooted in energy conservation and efficiency in mechanical systems.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done against gravity and the kinetic energy of the crate. There is a focus on understanding how to combine these energy components to find the total work output and subsequently the heat input.

Discussion Status

Some participants have provided guidance on correcting the initial calculations and emphasized the importance of including both gravitational potential energy and kinetic energy in the work done. Multiple interpretations of how to approach the problem are being explored.

Contextual Notes

Participants are questioning the significance of velocity in the context of work done and are clarifying the relationship between work, efficiency, and heat input. There is an acknowledgment of potential errors in the initial calculations.

mikefitz
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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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