# Help with the work-energy principle

• TiernanW
In summary, the conversation discusses a question about finding the displacement of an object using equations for potential energy, work done, and kinetic energy. The main point of confusion is the direction of the displacement and force vectors, which affects the sign of the final equation. The conversation concludes with a consensus that the displacement should be positive, and therefore, the final equation should be d = 3x/μ.
TiernanW

PE at A = 3mgx
WD = Fs
KE = 1/2mv^2

## The Attempt at a Solution

The question I am stuck on is part ii.

I worked out from part i that the PE at A is 3mgx, so therefore all this must go towards the KE and the sound, and doing work against friction, etc...

I said that the change in energy must equal the work done so -3mgx = μmgd, and the mg must cancel so d = -3x/μ

However the mark scheme says d = 3x/μ, without the negative. So have I done something wrong or is this just to do with the direction I have chosen?

The right side of the expression is negative because the displacement vector and the force vector are in opposite directions

rpthomps said:
The right side of the expression is negative because the displacement vector and the force vector are in opposite directions
The mark scheme has said: 3mgx = μmgd.

So if I had taken the directions the other way I would have said -3mgx = -μmgd?

The way I like to think of it is the loss (negative) in potential energy is equal to the loss of energy due to heat (friction). Both are negative values but algebraically they can be simplified to be positive.

rpthomps said:
The way I like to think of it is the loss (negative) in potential energy is equal to the loss of energy due to heat (friction). Both are negative values but algebraically they can be simplified to be positive.
How is that possible? Doesn't some of the energy go towards movement and sound etc?

Merely relocating something from X to Y takes no nett energy. The body starts from rest and ends up at rest so gains no KE, and in the absence of force fields it gains/loses no PE, and in the absence of friction it produces no heat (and no sound).

Certainly, to get it initially to begin moving you must give it energy, but as it nears destination Y you must arrange to take back all that kinetic energy so it comes to a standstill at Y. You get back the energy you put in, so the relocation itself has consumed zero nett energy.

NascentOxygen said:
Merely relocating something from X to Y takes no nett energy. The body starts from rest and ends up at rest so gains no KE, and in the absence of force fields it gains/loses no PE, and in the absence of friction it produces no heat (and no sound).

Certainly, to get it initially to begin moving you must give it energy, but as it nears destination Y you must arrange to take back all that kinetic energy so it comes to a standstill at Y. You get back the energy you put in, so the relocation itself has consumed zero nett energy.

Surface CD is rough so there is friction.

TiernanW said:
Surface CD is rough so there is friction.
That's the method provided here for taking back the energy. Once all KE has been removed by friction, the body is at standstill.

NascentOxygen said:
That's the method provided here for taking back the energy. Once all KE has been removed by friction, the body is at standstill.
So WD against friction = μmgd. Therefore since all PE is lost to this then 3mgx = μmgd, so d = 3x/μ. That makes a little more sense right now.

But what I'm trying to get my head around is that fact that ΔE = WD, so I would have thought work done is actually -3mgx, so -3mgx = μmgd? Forgive me if I am being slow. I just see that it has lost energy so the change is negative.

A positive force produces a positive acceleration, (F=m·a), so whichever direction you choose to consider positive for the friction force must likewise be the direction you use for positive displacement.

TiernanW
NascentOxygen said:
A positive force produces a positive acceleration, (F=m·a), so whichever direction you choose to consider positive for the friction force must likewise be the direction you use for positive displacement.
I understand it now. -3mgx = -μmg * d as friction acts in the opposite direction. :)

## 1. What is the work-energy principle?

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. In other words, the amount of work done on an object is directly proportional to the change in its speed or velocity.

## 2. How is the work-energy principle used in science?

The work-energy principle is used to explain the relationship between work, energy, and motion. It is often used in physics and engineering to analyze the behavior of objects and systems in motion.

## 3. What are the units of measurement for work and energy?

The units of measurement for work are joules (J) or newton-meters (N*m), while the units for energy are also joules (J). This is because work and energy are directly related to each other.

## 4. How is the work-energy principle related to other principles in physics?

The work-energy principle is closely related to the conservation of energy and the laws of motion. It can also be used in conjunction with other principles, such as the law of conservation of momentum, to analyze and understand the behavior of systems in motion.

## 5. How can the work-energy principle be applied to real-life situations?

The work-energy principle can be applied to various real-life situations, such as calculating the amount of work needed to lift an object, understanding the energy efficiency of different machines, and analyzing the motion of vehicles and projectiles. It is a fundamental principle that can be applied to many different scenarios in our daily lives.

• Introductory Physics Homework Help
Replies
7
Views
991
• Introductory Physics Homework Help
Replies
8
Views
893
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
58
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
29
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
685
• Introductory Physics Homework Help
Replies
33
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
315