Are the energy requirements the same for moving masses in different scenarios?

  • I
  • Thread starter Fizzics
  • Start date
  • Tags
    Comparison
In summary, the conversation discusses two scenarios for comparison. Scenario 1 involves moving two connected masses 3 meters apart on a friction-free surface using the same force and acceleration in a straight line, while Scenario 2 involves moving the same masses over an elevated friction-free pulley. The question is whether the energy required to move Scenario 1 is the same as in Scenario 2, and if the two masses moving over a pulley can be considered the same as the two masses moving in a straight line. The conversation also touches on the concept of conservation of energy and how it simplifies the problem. The solution to the question is to compute the energy of each mass before and after the experiment, and subtract the initial energy from the final energy
  • #1
Fizzics
26
0
Here there are two scenarios for comparison.

Scenario 1: We move two connected masses which are Constantly 3 metres apart on a friction free surface using the same force and acceleration in a straight line.

Scenario 2: We move two connected masses which are Constantly 3 metres apart over an elevated friction free pulley using the same force and acceleration as above.

Question: Is the energy required to move scenario 1 the same in scenario 2? and can the 2 masses moving over a pulley be considered the same as the 2 masses moving in a straight line?
 
Physics news on Phys.org
  • #2
Fizzics said:
Here there are two scenarios for comparison.

Scenario 1: We move two connected masses which are Constantly 3 metres apart on a friction free surface using the same force and acceleration in a straight line.

Scenario 2: We move two connected masses which are Constantly 3 metres apart over an elevated friction free pulley using the same force and acceleration as above.

Question: Is the energy required to move scenario 1 the same in scenario 2? and can the 2 masses moving over a pulley be considered the same as the 2 masses moving in a straight line?
Over a pulley implies they are being raised against the force of gravity, versus your horizontal frictionless table. Is that really the comparison you are asking about?
 
  • #3
berkeman said:
Over a pulley implies they are being raised against the force of gravity, versus your horizontal frictionless table. Is that really the comparison you are asking about?
As one mass is going up the other mass is going down the other side of the pulley
 
  • #4
Mass movement picture.png
 
  • #5
Fizzics said:
As one mass is going up the other mass is going down the other side of the pulley
Then how are they always 3 meters apart and connected? Sorry that I'm not getting the setup for the question.
 
  • #6
Oh, I see the figure now. You mean "along the rope" a constant distance apart.
 
  • #7
berkeman said:
Oh, I see the figure now. You mean "along the rope" a constant distance apart.
That's right
 
  • #8
Can you draw free body diagrams (FBDs) for each situation? That may help you to figure out the answer.

Also, does it make any difference if a force F is applied left to the leftmost mass in the upper diagram and down on the left mass in the pulley diagram, versus applying F/2 to each mass?

:smile:
 
  • #9
You can solve this problem by simply comparing the final energies (potential energy proportional to height above the ground, and kinetic energy proportional to the square of speed) with the initial energies.

What happens between initial and final, the masses go up/down sideways or in circles, doesn't matter.
 
  • #10
berkeman said:
Can you draw free body diagrams (FBDs) for each situation? That may help you to figure out the answer.

Also, does it make any difference if a force F is applied left to the leftmost mass in the upper diagram and down on the left mass in the pulley diagram, versus applying F/2 to each mass?

:smile:
The equal forces would be applied to mass A sideways and Mass A downwards
 
  • #11
anorlunda said:
You can solve this problem by simply comparing the final energies (potential energy proportional to height above the ground, and kinetic energy proportional to the square of speed) with the initial energies.

What happens between initial and final, the masses go up/down sideways or in circles, doesn't matter.
Sorry but I don't understand what you are saying, Maybe I should say that the initial force is applied to Mass A in both diagrams and mass B follows because it is connected by a belt.
 
  • #12
Your question was how much energy does it take to ...

Conservation of energy is a wonderful principle. It permits huge simplifications of problems like this. Conservation of energy gives the answer to your question as:

##E_{net.added}=E_{final}-E_{initial}##

The details of how and when the energy was added, are irrelevant. All forces and directions of forces are irrelevant. No matter how complex the problem, conservation of energy requires that the answer is reduced to that simple equation. That's why I say that it simplifies.
 
  • Like
Likes berkeman
  • #13
anorlunda said:
Your question was how much energy does it take to ...

Conservation of energy is a wonderful principle. It permits huge simplifications of problems like this. Conservation of energy gives the answer to your question as:

##E_{net.added}=E_{final}-E_{initial}##

The details of how and when the energy was added, are irrelevant. All forces and directions of forces are irrelevant. No matter how complex the problem, conservation of energy requires that the answer is reduced to that simple equation. That's why I say that it simplifies.
So if it took 50 N initially to move the mass and 45 N Finally to stop the mass the net loss would be 5 N in both cases, is that correct?
 
  • #14
Fizzics said:
So if it took 50 N initially to move the mass and 45 N Finally to stop the mass the net loss would be 5 N in both cases, is that correct?
Energy is not measured in N.
 
  • #15
Fizzics said:
So if it took 50 N initially to move the mass and 45 N Finally to stop the mass the net loss would be 5 N in both cases, is that correct?

No, forces don't enter into the problem at all if you use energies.

The energy of each mass is ##\frac{1}{2}m*v^2 + m*g*h## where m is the mass, v is the velocity, g is 9.81 in Earth's gravity. and h is the height above the ground. Compute that for mass 1 and mass 2 both before the experiment starts and after it ends. Subtract initial from final and you have the energy that was put into the system. If the starting or ending velocities are zero, then v=0. You can also choose a reference where h=0 for one of the starting positions.

Be careful with the units if you want the correct answer.
 
  • #16
Fizzics said:
So if it took 50 N initially to move the mass and 45 N Finally to stop the mass the net loss would be 5 N in both cases, is that correct?

There is another possibility. You original question asked how much energy. Did you mean to say how much force?
 
  • #17
anorlunda said:
There is another possibility. You original question asked how much energy. Did you mean to say how much force?
I guess I have wandered between force and energy without thinking so to try and clarify let's stick with energy required. So let's say that 100 joules of energy is applied to moving mass A in both scenarios in the direction of the arrows for 2 seconds (= 200 watts) then as a result of this let's say that the straight line masses in scenario 1 move for 2 metres before coming to rest, my question is would the masses moving over the pulley in scenario 2 also move for 2 metres before coming to rest. Thank you in advance for taking the time to consider this.
 
  • #18
Fizzics said:
that 100 joules of energy is applied to moving mass A in both scenarios in the direction of the arrows for 2 seconds (= 200 watts)
200 watts for 2 seconds is 400 joules, not 100 joules

Fizzics said:
then as a result of this let's say that the straight line masses in scenario 1 move for 2 metres before coming to rest, my question is would the masses moving over the pulley in scenario 2 also move for 2 metres before coming to rest.
What makes them come to rest?
 
  • #19
A.T. said:
200 watts for 2 seconds is 400 joules, not 100 joulesWhat makes them come to rest?
They come to rest naturally because of the forces around them such as gravity and air resistance, they are not in a vacuum or perpetual motion. The 2m stopping time is just so that a comparison can be made between the 2 scenarios it could just as easily be a 2 km stopping distance providing there was enough distance between the masses going over the pulley. I guess that my real question is that does gravity have more of an effect in stopping the masses in either of the scenarios because of the way they are set up?
 
  • #20
A.T. said:
200 watts for 2 seconds is 400 joules, not 100 joulesWhat makes them come to rest?
My meaning was 100 J per second for 2 seconds = total energy used 200w (1 Watt is the power of a Joule of energy per second ) however this was only an example and not relevant to my questions outcome
 
  • #21
Fizzics said:
They come to rest naturally because of the forces around them such as gravity and air resistance,
Same air resistance at same speed in both scenarios?

Fizzics said:
I guess that my real question is that does gravity have more of an effect in stopping the masses in either of the scenarios because of the way they are set up?
None in the first case. And none in the second if the masses are equal.

Fizzics said:
total energy used 200w
Energy is measured in Joule, not Watt.
 
  • #22
Thank you
So two equal masses traveling over a pulley is the same as two equal masses traveling in a straight line if all of the conditions are the same?
 
  • #23
Fizzics said:
My meaning was 100 J per second for 2 seconds = total energy used 200w (1 Watt is the power of a Joule of energy per second ) however this was only an example and not relevant to my questions outcome

You're getting it backward. 100 watts for 2 seconds = 200 watt seconds = 200 joules

Fizzics said:
They come to rest naturally because of the forces around them such as gravity and air resistance,
Then your problem is misstated. You must either assume that the objects are in empty space, or you must explicitly include gravity and friction in the calculation. The devil is in the details. If you get the details wrong, it confuses you about the principles.
 
  • #24
anorlunda said:
You're getting it backward. 100 watts for 2 seconds = 200 watt seconds = 200 joules Then your problem is misstated. You must either assume that the objects are in empty space, or you must explicitly include gravity and friction in the calculation. The devil is in the details. If you get the details wrong, it confuses you about the principles.
Thanks for the feedback I will take it all on board
 

1. What is a comparison of moving masses?

A comparison of moving masses is a scientific experiment that involves measuring the differences in motion between two or more objects with different masses. It is used to study the effects of mass on an object's acceleration and velocity.

2. How is a comparison of moving masses conducted?

A comparison of moving masses is typically conducted by setting up a track with a pulley system and using carts of different masses. The carts are then released from the same height and their motion is recorded and analyzed.

3. What are the key findings from a comparison of moving masses?

The key findings from a comparison of moving masses include the fact that objects with larger masses have a greater resistance to changes in motion and require more force to accelerate, while objects with smaller masses experience greater acceleration.

4. What is the significance of a comparison of moving masses in physics?

A comparison of moving masses is significant in physics because it helps to illustrate the fundamental principles of Newton's laws of motion and the concept of inertia. It also allows for the calculation of important variables such as force and acceleration.

5. How can a comparison of moving masses be applied in real-life situations?

A comparison of moving masses can be applied in real-life situations to understand and improve the performance of various machines and vehicles, such as cars and airplanes. It can also help in designing safer amusement park rides and sports equipment.

Similar threads

Replies
51
Views
1K
  • Special and General Relativity
Replies
33
Views
1K
  • Mechanical Engineering
Replies
2
Views
664
  • Special and General Relativity
Replies
9
Views
761
  • Special and General Relativity
2
Replies
36
Views
632
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
376
  • Introductory Physics Homework Help
Replies
29
Views
929
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
1K
Back
Top