Energy required to lift walls

  • Thread starter Thickmax
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  • #1
Thickmax
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Homework Statement:
Pre University qualification top up course
Relevant Equations:
See below
1624908585995.png


Please help, I'm really struggling to understand how to work this...

work = mass * gravity * change in height

w = m*g*(h2 - h1)

Total energy = Sum of potential energy + Kinetic Energy

work = 1/2Kinetic Energy * Potential energy ^2

w = (1/2 * K) * U^2

m*g*(h2-h1)=(1/2*k)*U^2

rearanged = (2(m*g*(h2-h1))/U^2)=K

We are given

U, m, g, h2 and h1


Firstly, is my above working along the right line.

Secondly, how to I show this as 80%?

Please help
 

Answers and Replies

  • #2
Doc Al
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work = mass * gravity * change in height

w = m*g*(h2 - h1)
Keep it simple. That's the work done -- and thus the energy expended -- to lift one wall. So, how much energy is needed to lift N walls? Set that equal to 80% of the battery's energy to see how many can be lifted.

Don't worry about KE. Just assume the walls are lifted without any appreciable increase in KE.
 
Last edited:
  • #3
Thickmax
31
8
I've never seen that equation before. What are the units of both sides? :smile:

Imagine that you lift each wall so slowly that it's kinetic energy can always be taken as zero. The work done by the battery to raise a single wall is ##mg(h_2-h_1)##, as you wrote. How much work does the battery then do to raise ##N## walls; and to what fraction of the number ##U## does this correspond?
If you've never seen it before, I've probably not done it right!

W=F*Displacement

Integrated = K*potential energy

Integrated = 1/2*K*potential energy^2
 
  • #4
Doc Al
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W=F*Displacement
That makes sense. It's the definition of work.

But the following does not make sense. What are you integrating?
Integrated = K*potential energy

Integrated = 1/2*K*potential energy^2
 
  • #5
Thickmax
31
8
Keep it simple. That's the work done -- and thus the energy expended -- to lift one wall. So, how much energy is needed to lift N walls? Set that equal to 80% of the battery's energy to see how many can be lifted.

Don't worry about KE. Just assume the walls are lifted without any appreciable increase in KE.
ok, so N(m*g*(h2 - h1))=U?

and I need to rearrange (N(m*g*(h2-h1))=U

N=U/(m*g*(h2-h1)) right?
 
  • #6
Doc Al
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ok, so N(m*g*(h2 - h1))=U?

and I need to rearrange (N(m*g*(h2-h1))=U

N=U/(m*g*(h2-h1)) right?
Almost. Don't set the energy needed equal to U. U is the total capacity of the battery -- you only want to use a fraction of that total.
 
  • #7
Thickmax
31
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Ok, so...

(N(m*g*(h2-h1))=4/5U

N=4/5U/(m*g*(h2-h1))
 
  • #8
Doc Al
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Now you've got it.
 
  • #9
Thickmax
31
8
Now you've got it.
Mind blown how it makes sense now!

Thank you soo much for your help
 

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