PERES LAWHow does Ampere's Law relate to electromagnetism?

[mikefitz]

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

 

[Andrew Mason]

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

 

[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?

 

[Andrew Mason]

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