Conservation of Energy Homework problem

In summary: Wext = ΔKE but the external work is just mgΔh so then would Wext = ΔKE = mgΔh?Yes, that is correct. In summary, the net work on system T is equal to the change in kinetic energy, which can be calculated using the formula ΔKE = 1/2mvf^2. The change in kinetic energy is also equal to the external work done on the system, which can be expressed as Wext = mgΔh. Therefore, Wext = ΔKE = mgΔh. The change in potential energy for system T is 0 since the only force acting on the ball is weight and the ball is not moving vertically.
  • #1
jlmccart03
175
9

Homework Statement


A ball of mass m falls from height hi to height hf near the surface of the Earth. When the ball passes hf, it has a speed of vf. Ignore air resistance. Consider the system T which consists of the ball only.

Write an expression for each of the following quantities in terms of the given variables and any physical constants:
The net work on system T
The change in kinetic energy of system T
The change in potential energy of system T

Homework Equations


ΔKE = 1/2mv2
ΔPE = -mgh = -mg(hf - hi)
Wnet ext = ΔKE

The Attempt at a Solution


I know that the total energy throughout the system does not change. However, Potential Energy caused by gravity does change into Kinetic Energy as the ball falls for the initial height. Therefore I believe that the net work on system T is ΔKE = 1/2mvf2.

The change in kinetic energy is ΔKE = 1/2mvf2

And the change in Potential energy is 0.

This is where I felt that I may be completely wrong since the net work is all the forces multiplied by displacement I think my net work formula or "expression" is incorrect. Could anyone help me verify my mistake? Thanks!
 
Physics news on Phys.org
  • #2
Hi!

First of -
Could you list all the forces acting on the ball?

Now,
jlmccart03 said:
Wnet ext = ΔKE
The expression you have used here is the Work-Energy theorem. What is the fundamental definition of work? What about the potential energy?
 
  • Like
Likes jlmccart03
  • #3
It's worth pointing out two things about your approach so far:
jlmccart03 said:

Homework Equations


ΔKE = 1/2mv2
ΔPE = -mgh = -mg(hf - hi)
Wnet ext = ΔKE
Do you see that your ΔKE expression implies an initial state of rest (if what you mean by v is vf) ? You did not state that as an assumption, so you might want to make sure that it's an assumption worth making.

jlmccart03 said:

The Attempt at a Solution


I know that the total energy throughout the system does not change. However, Potential Energy caused by gravity does change into Kinetic Energy as the ball falls for the initial height.
...
And the change in Potential energy is 0.

Either the PE changes by some nonzero amount or it doesn't change. You can't have both.
 
  • Like
Likes jlmccart03
  • #4
LemmeThink said:
Hi!

First of -
Could you list all the forces acting on the ball?

Now,

The expression you have used here is the Work-Energy theorem. What is the fundamental definition of work? What about the potential energy?

Okay, the only force acting on the ball in this scenario is weight or mg.

The fundamental definition of work is W = F * d or force times displacement. Potential energy due to gravity (Since technically we are dealing with PEgrav) is defined as PE = m*g*h or mass times gravity times height.

So I am assuming that the work done on the system T would be mgΔh as indicated by the definition of work?
 
Last edited:
  • #5
jlmccart03 said:
Okay, the only force acting on the ball in this scenario is weight or mg.

The fundamental definition of work is W = F * d or force times displacement. Potential energy due to gravity (Since technically we are dealing with PEgrav) is defined as PE = m*g*h or mass times gravity times height.
That is correct.
jlmccart03 said:
So I am assuming that the work done on the system T would be -mgΔh as indicated by PE's formula?
No. Work done, would be, as you correctly stated, ##\vec F.\vec d##, along with the appropriate sign. You should not be using PE for determining the work done. While you could make use of the potential energy, it is better to work with the definition ##W = \vec F.\vec d## So tell me, is gravity doing positive, or negative work?
 
  • Like
Likes jlmccart03
  • #6
LemmeThink said:
That is correct.

No. Work done, would be, as you correctly stated, ##\vec F.\vec d##, along with the appropriate sign. You should not be using PE for determining the work done. While you could make use of the potential energy, it is better to work with the definition ##W = \vec F.\vec d##

I realized what I said originally was wrong and fixed it within my reply.
LemmeThink said:
So tell me, is gravity doing positive, or negative work?

It is most definitely doing positive work since the motion of the ball and gravity are in the same direction correct?
 
  • #7
Yes; and it's value?
 
  • Like
Likes jlmccart03
  • #8
jlmccart03 said:
Okay, the only force acting on the ball in this scenario is weight or mg.

The fundamental definition of work is W = F * d or force times displacement. Potential energy due to gravity (Since technically we are dealing with PEgrav) is defined as PE = m*g*h or mass times gravity times height.

So I am assuming that the work done on the system T would be -mgΔh as indicated by PE's formula and the definition of work?

If you consider the positive direction for the motion to be "up", and the ball is falling "down", then the work could be expressed as -mgΔh as long as we consider the
Δh to be negative due to the choice of orientation for our coordinate system.
However, like what LemmeThink said, you don't really need to consider potential energy in your approach here. The force of weight being -mg, and the displacement d = Δh gives you the work using the fundamental definition. From this expression for work we can come to our definition of ΔPE as the negative of the work done by gravity.
 
  • Like
Likes jlmccart03
  • #9
LemmeThink said:
Yes; and it's value?
It would be mgΔh correct? Since the force is mg and d is the height h.
 
  • Like
Likes LemmeThink
  • #10
Tallus Bryne said:
If you consider the positive direction for the motion to be "up", and the ball is falling "down", then the work could be expressed as -mgΔh as long as we consider the
Δh to be negative due to the choice of orientation for our coordinate system.
However, like what LemmeThink said, you don't really need to consider potential energy in your approach here. The force of weight being -mg, and the displacement d = Δh gives you the work using the fundamental definition. From this expression for work we can come to our definition of ΔPE as the negative of the work done by gravity.
Oh, okay, I think I get it now. From the work expression I can then determine my energy formulas (or at least my potential energy formula) since the net Work is equal to these two formulas or am I still way off from the target?
 
  • #11
jlmccart03 said:
Oh, okay, I think I get it now. From the work expression I can then determine my energy formulas (or at least my potential energy formula) since the net Work is equal to these two formulas or am I still way off from the target?

We're dealing with gravity, but the following approach is supposed to work with any force that can be considered conservative (means that work done by that force is path independent). You derive an expression for the work done by that force over some specified path, which begins at point A and ends at point B, or initial and final heights, for example. The result can be used as a basis for defining a potential energy for that force, equating the negative of the work done by the force to the change in that potential energy.
So instead of using the work-kinetic energy theorem as a starting point in a problem, where the net work expression involves all the forces acting on the object, you can start with a statement of the conservation of energy: ΔKE + ΔPE = W , where W is now just the work done by all the non-conservative forces (like friction as a common example in intro physics problems), and the work expressions for all conservative forces are moved to the "energy side" of the equation.
 
  • Like
Likes jlmccart03
  • #12
Tallus Bryne said:
We're dealing with gravity, but the following approach is supposed to work with any force that can be considered conservative (means that work done by that force is path independent). You derive an expression for the work done by that force over some specified path, which begins at point A and ends at point B, or initial and final heights, for example. The result can be used as a basis for defining a potential energy for that force, equating the negative of the work done by the force to the change in that potential energy.
So instead of using the work-kinetic energy theorem as a starting point in a problem, where the net work expression involves all the forces acting on the object, you can start with a statement of the conservation of energy: ΔKE + ΔPE = W , where W is now just the work done by all the non-conservative forces (like friction as a common example in intro physics problems), and the work expressions for all conservative forces are moved to the "energy side" of the equation.
Okay so, the net work done on the system is mgΔh, ΔKE = 1/2mvf2 and PE = -mgΔh in this scenario correct? This is based on the definition of work and definition of PE.
 
  • #13
jlmccart03 said:
Okay so, the net work done on the system is mgΔh, ΔKE = 1/2mvf2 and PE = -mgΔh in this scenario correct? This is based on the definition of work and definition of PE.

This is fine if you treat Δh as a positive number (this would work with a coordinate system with the downward direction being positive). Perhaps I'm being a little too pedantic about consistency with signs and coordinate systems. The most important thing is that your result for work is positive, ΔKE is equal in value but opposite in sign to ΔPE (no matter what your coordinate system). So in your case that makes ΔKE positive and ΔPE negative by the same amount.
 

1. What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

2. How do you apply the law of conservation of energy to a homework problem?

To apply the law of conservation of energy to a homework problem, you must identify all the different forms of energy involved in the system and track their transformations. This can be done using equations such as the energy balance equation or the work-energy theorem.

3. What are some common types of energy involved in conservation of energy problems?

Some common types of energy involved in conservation of energy problems include kinetic energy, potential energy, thermal energy, electrical energy, and chemical energy.

4. Can energy ever be completely conserved in real-world situations?

No, energy can never be completely conserved in real-world situations. This is due to factors such as friction and energy losses in a system, which result in some of the energy being converted into forms that are not useful for the intended purpose.

5. How can I check if my conservation of energy calculations are correct?

To check if your conservation of energy calculations are correct, you can perform a dimensional analysis to ensure that the units on both sides of the equation match. You can also compare your results to known values or use a different method to solve the problem and see if you get the same answer.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
733
Replies
10
Views
313
  • Introductory Physics Homework Help
Replies
15
Views
300
  • Introductory Physics Homework Help
2
Replies
55
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
935
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
366
  • Introductory Physics Homework Help
Replies
24
Views
891
Replies
33
Views
3K
Back
Top