-Change in Mechanical Energy (Gr 11 Prac)-

[Dovekie]

Greetings all I'm new to this forum, so please spare me if I make a fool of myself in this post.

I was working on a Physics Practical this morning at school concerning Change in Mechanical Energy, in relation to Potential and Kinetic Energy. Although it seems like a easy task, I'm getting confused by all the calculations.

Outline of the prac:
We had a wooden board about 1.79 metres in length and we had to slide a wooden block (232.3 grams) from the top of the board to the bottom, measuring the time it took to slide from the top to the bottom. We repeated the same steps but with the wooden board placed at varying heights. (E.g. 0.7m, 0.85m, 1m, 1.15m, 1.30m), therefore also at different angles (which I've managed to calculate already). Then by using our results (the time it took for the block to reach the ground for each height) we have to calculate the Change in Mechnical Energy.

There are several different ways to calculate ME, and I'm confused about which one is actually the correct procedure. If anyone has the time to give me some suggestions, it would be greatly appreciated.

The formula for Mechanical Energy is: ME = (1/2)mv^2 + mgh
(ME = Kinetic Energy + Potential Energy), but what numbers do I use to work out KE and PE?

Thanks.

 

[HallsofIvy]

Did you think to measure the height of the upper end of the board (above the lower end) at each angle? If you didn't you can calculate it by : length of board* sin(angle).
That height, h, in mgh, gives you the potential energy at the top of the board (and so the total energy just as it was starting to slide). At the bottom of the board, the potential energy is 0. IF there were no friction, so that energy was conserved, the potential energy at the top (the total energy) would have been converted to kinetic energy and you could use that to calculate the speed at the bottom.

In this experiment, you calculated the time the block took to slide the length of the board so you can calculate its (average) speed. As long as the angle was not too steep, the block slides at a constant speed so the speed you calculate can be taken as the speed at the end- you can calculate kinetic energy (= total energy at bottom) from (1/2)mv².

You should find that the total energy (kinetic energy) at the bottom is lower than the total energy (potential energy) at the top- that's the "change in mechanical energy": there is friction so mechanical energy is not conserved. (Total energy is still conserved- the friction causes both the block and board to become a little hotter.)

 

[Dovekie]

Friction was neglected in this experiment =)

And I actually forgot the measure the angle, but it wouldn't have been very accurate anyway with a smalle protractor anyhow, so I resorted to using the Sine rule. I think the purpose of the prac was to actually calculate whether the block accelerated as it slid down the ramp, but I worked out the acceration using v(final) = v(initial) + at -- which I hope gave the correct acceleration. v(initial) I assumed was 0 because the block started form rest at the top of the wooden board.

We were told that PE + KE is always a constant number in a set experiment. I.E. If PE at one stage is 40, and KE at the same stage is 50 = and when added together is 90. Then when PE at another stage is 70, KE would have to be 20 in order for the total ME to remain constant at 90. We were told something along those lines, so I'm still kind of confused. If ME = PE + KE and we have the find the change, but it's always constant... there would be no change... either that or I'm completely off track, which I think is very likely.

Nevertheless, you were still a great help! Thanks much =)

 

[Nenad]

We were told that PE + KE is always a constant number in a set experiment. I.E. If PE at one stage is 40, and KE at the same stage is 50 = and when added together is 90. Then when PE at another stage is 70, KE would have to be 20 in order for the total ME to remain constant at 90. We were told something along those lines, so I'm still kind of confused. If ME = PE + KE and we have the find the change, but it's always constant... there would be no change... either that or I'm completely off track, which I think is very likely.

there is something that doesn't make sense in you experiment. ME = PE + KE, but since you are starting from rest in all of your trials, then there would be no need for this. You would use PE = KE (mgh = 1/2mv^2), and caculate that out. I think your teacher wanted you to calculate the cahnge in potential energy and kinetic energy. The kinetic would be less because of friction. That is the only reason I see for the experiment, so see the effects of friction on a ramp at different slopes. Hope I've been a help.
oh and PE = mgh, m=mass, g=9.8N/kg, h=vertical height
KE = 1/2mv^2, m=mass, v=speed in m/s

 

[JohnDubYa]

We were told that PE + KE is always a constant number in a set experiment. I.E. If PE at one stage is 40, and KE at the same stage is 50 = and when added together is 90. Then when PE at another stage is 70, KE would have to be 20 in order for the total ME to remain constant at 90. We were told something along those lines, so I'm still kind of confused. If ME = PE + KE and we have the find the change, but it's always constant... there would be no change... either that or I'm completely off track, which I think is very likely.

You are completely off track because you have been misinformed.

It is not true that in any set (?) experiment that PE + KE is constant. To understand how energies and forces interact, you need to learn and understand the work-energy theorems:

[tex] W_{\rm nc} = \Delta({\rm ME})[/tex],

that is, the work done by a non-conservative force on an object changes the total mechanical energy of the object by an amount [tex]\Delta({\rm ME})[/tex],

[tex] W_{\rm c} = -\Delta({\rm PE})[/tex],

the work done by a conservative force on an object changes the potential energy of the object by an amount [tex]-\Delta({\rm PE})[/tex],

For PE + KE to remain constant, no non-conservative force can do work on it. However, friction (which is inherently non-conservative) does do work on the block in your experiment, so the total mechanical energy does change. Friction is NOT negligible in your experiment. If it was, the answer to your question about the change in mechanical energy would always be 0. ( Why? Look at the work-energy theorems.)

 

[maverick280857]

A problem with some of our Newtonian mechanics courses is the non-correlation of conservative and nonconservative forces. The terms come up only during the study of a topic called work and energy. Even while studying work and energy, the two methods of analysis, namely the Kinetic Energy-Work Theorem and the "Principle of Conservation of Energy" remain isolated from each other with regard to the consideration of nonconservative forces.

In physics, you need to understand the effect of nonidealities in nature and friction is indeed one of them (though it does aid us a lot really :-D).

Another way to express the mechanical energy equation is to write

[tex]
\Delta V_{g} + \Delta V_{e} + \Delta T = \Delta U
[/tex]

where [tex]\Delta V_{g}[/tex] represents the change in gravitational potential energy, [tex]\Delta V_{e}[/tex] represents the change in elastic potential energy, [tex]\Delta T[/tex] is the change in kinetic energy and [tex]\Delta U[/tex] is the work done by forces other than gravitational and elastic (spring like) forces.

In your case, [tex]\Delta U[/tex] represents the work done by friction and since no spring force is involved, you can drop the second term.

I can rewrite this equation as

[tex]
E_{final} = E_{initial} + W_{nc}
[/tex]

where [tex]W_{nc}[/tex] is the work done by a nonconservative force (this will indeed tend to decrease the initial energy). This equations seems more obvious as you understand how the work done by a nonconversative force changes the rather nice looking expression if it were absent,

[tex]
E_{final} = E_{initial}
[/tex]

Hope that helps...

Cheers
Vivek

 

[Dovekie]

Thank you all for your help, you put me on the right track. I finally understand my prac!

Thanks much.

 

[JohnDubYa]

(this will indeed tend to decrease the initial energy)

Not necessarily. For example, if I lift an object at constant speed, I am increasing the object's total mechanical energy.

 

[maverick280857]

Not necessarily. For example, if I lift an object at constant speed, I am increasing the object's total mechanical energy.

I said so about the nonconservative term in this case (such as friction). When you raise a body, you are doing work against a conservative field (the gravitational field). You are increasing the body's potential energy. But where is the nonconservative force acting on it? In fact for such a case, the [tex]\Delta U[/tex] will be zero in my equation (see below). Alternatively, set [tex]W_{nc}[/tex] equal to zero in my second equation.

Cheers
Vivek

 

[JohnDubYa]

The normal force you apply to the objectis a non-conservative force.


Remove gravity altogether. You apply the same normal force and, since there is no other force to counteract, accelerate the object. Again, you end up increasing the object's total mechanical energy (in the form of kinetic energy instead of potential energy).

 

[maverick280857]

The normal force you apply to the objectis a non-conservative force.


Remove gravity altogether. You apply the same normal force and, since there is no other force to counteract, accelerate the object. Again, you end up increasing the object's total mechanical energy (in the form of kinetic energy instead of potential energy).

John, my answer referred to your earlier statement, "Not necessarily. For example, if I lift an object at constant speed, I am increasing the object's total mechanical energy." which I thought you made about a body being lifted up only under the influence of gravity. You can effect this process of lifting up a body at constant speed only if you visualize the process as a quasistatic process...

Anyway, I do understand the point you made in your recent post and I quite agree with it.

Cheers
Vivek

 

[JohnDubYa]

John, my answer referred to your earlier statement, "Not necessarily. For example, if I lift an object at constant speed, I am increasing the object's total mechanical energy." which I thought you made about a body being lifted up only under the influence of gravity. You can effect this process of lifting up a body at constant speed only if you visualize the process as a quasistatic process...

When I used the word "lift," it should have been apparent that I was applying a normal force to the object, especially since I stated that the object was moving at constant speed. I am not sure what confused you.

In fact for such a case, the [tex]\Delta U[/tex] will be zero in my equation (see below). Alternatively, set equal to zero in my second equation.

I am not sure I follow. If a body is rising, its potential energy increases; it does not remain a constant.

 

[maverick280857]

John, please read my first post again. I clearly defined [tex]\Delta U[/tex] as the work done by forces other than elastic, gravitational forces.

 

[JohnDubYa]

Sorry for the mistake, but you are using some very unfortunate notation since U usually refers to potential energy in general. It would have helped if you had affixed a prime to the letter.

 

[maverick280857]

Well the notation I mentioned is a nonstandard one I agree, so forgive me for any confusion caused by it :-) You could use something else (say [tex]W_{other}[/tex]) to denote the work done by all forces other than elastic and gravitational. Anyway, the notation makes no difference whatsoever and is purely your choice.

Cheers
Vivek