Understanding the Energy Required to Lift Walls

In summary, the equation states that if you lift a wall with a battery, it will expend 4/5 of the battery's energy.
  • #1
Thickmax
31
8
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 h1Firstly, is my above working along the right line.

Secondly, how to I show this as 80%?

Please help
 
Physics news on Phys.org
  • #2
Thickmax said:
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:
  • Like
Likes Thickmax
  • #3
ergospherical said:
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
Thickmax said:
W=F*Displacement
That makes sense. It's the definition of work.

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

Integrated = 1/2*K*potential energy^2
 
  • #5
Doc Al said:
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
Thickmax said:
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
Ok, so...

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

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

Thank you soo much for your help
 

What is the definition of energy required to lift walls?

The energy required to lift walls refers to the amount of energy needed to move a wall from its initial position to a higher position, against the force of gravity.

How is the energy required to lift walls calculated?

The energy required to lift walls can be calculated using the formula E = mgh, where E is the energy, m is the mass of the wall, g is the acceleration due to gravity, and h is the height the wall is lifted.

What factors affect the energy required to lift walls?

The energy required to lift walls is affected by the mass of the wall, the height it is lifted, and the force of gravity. Other factors such as friction and air resistance may also play a role.

Can the energy required to lift walls be reduced?

Yes, the energy required to lift walls can be reduced by using techniques such as pulleys or levers, which can decrease the force needed to lift the wall. Additionally, using lighter materials for the wall can also reduce the energy required.

Why is understanding the energy required to lift walls important?

Understanding the energy required to lift walls is important for engineers and architects in designing structures that are safe and efficient. It also helps in determining the amount of work and resources needed for construction projects.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
317
  • Introductory Physics Homework Help
Replies
5
Views
417
  • Introductory Physics Homework Help
Replies
14
Views
874
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
792
  • Introductory Physics Homework Help
Replies
4
Views
820
  • Introductory Physics Homework Help
Replies
13
Views
630
  • Introductory Physics Homework Help
Replies
3
Views
368
  • Introductory Physics Homework Help
Replies
3
Views
275
  • Introductory Physics Homework Help
Replies
17
Views
1K
Back
Top