The conservation of energy, why does it work? Negative work vs positive work

flyingpig

1. The problem statement, all variables and given/known data


This is the conservation of energy in the absence of external forces


ΔKE + ΔPE + ΔTE + ΔU= 0

TE is friction, U is other internal energy, like the spring

So expand and disregard U and TE for a moment


½m(v² - v₀²) + mg(h - h₀) = 0

Now comes the question, why is + mg(h - h₀) and not -mg(h - h₀)? Is gravitational potential energy not the same thing as the work done by gravity? Because gravity pulls down and if you go up, that's negative work. I've used the equation (with the positive mgh) many times and have not failed me.

PhanthomJay

Well you may have just been lucky . The work done by a conservative force, like gravity, is the negative of its potential energy change (W_g = - ΔPE). For example, if an object is dropped from rest from a building of height h, then ½m(v² - 0) + mg(0 - h) = 0, or
½m(v²) = mgh, solve for v, and note the change in potential energy is negative, and the work done by gravity is positive.

If, on the other hand, an object is thrown upward from the ground at an initial speed v, to a maximum height of h, then
½m(0 - v²)+ mg(h - 0) = 0, or
- ½m(v²) = -mgh, solve for v, and note the change in potential energy is positive, and the work done by gravity is negative. Gotta love those plus and minus signs...one of the more difficult concepts of Physics to grasp.

flyingpig

No, I am questioning about my formula, shouldn't it be -mg(h - h₀) instead of mg(h - h₀)?


½m(v² - v₀²) + mg(h - h₀) = 0 <==== this one works

or

½m(v² - v₀²) - mg(h - h₀) = 0 <==== this one does not, but it is correct

p21bass

Remember that g is -9.81m/s². Thus, when you plug it into your equation, positive mgh has a negative value.

flyingpig

My book did not use the -9.8m/s^2 and I thought we consider g only in its magnitude.

Doc Al

No. As a mass is raised, the gravitational PE increases. (Gravitational PE is minus the work done by gravity. It's the work done against gravity.)

No, g stands for the magnitude of the acceleration due to gravity; it's always positive. (Taking down as negative, the acceleration due to gravity is -g.)

p21bass

Oh, right. Oops... Sorry about that.

flyingpig

That just confused me even more. I have two Physics texts that gives me two different formulas, but they seem to work the same way except things are arranged on different sides of the equation.

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

Ex. Problem using this problem using the equation given above

A crate of mass m = 10.7 kg is pulled up a rough incline with an initial speed of vi = 1.59 m/s. The pulling force is F = 97 N parallel to the incline, which makes an angle of θ = 19.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.09 m.

(a) What is the change in kinetic energy of the crate?

Force done by gravity = -180J

Force done by friction also the change in internal energy = 201J

Force done on the crate = 494J

[tex]\Delta KE = \Sigma W_{other} - \Delta E_{int}[/tex]

=494J + (-180J) - 201J
= 113J

Now my question is how is that formula different from the one in my first post. Also why is internal energy considered as friction? If that is the case, what is spring potential energy? I thought that is internal energy

Abdul Quadeer

The total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do no work. This is the principle of conservation of energy.
So K + U = constant
Ki + Ui = Kf + Uf
You can see that ' ½m(v² - v₀²) + mg(h - h₀) = 0 ' is in accordance with the above equation. If you use - mg(h - h₀), it will violate the law.

Abdul Quadeer

The total mechanical energy K + U is not constant if non-conservative forces like friction etc act between the parts of the system. Hence we can't apply the principle of conservation of energy in presence of non-conservative forces.
However, the work-energy theorem is still valid. So according to it,

work done by all the forces equals the change in kinetic energy.
Wc + Wnc + Wext = Kf-Ki
where c=conservative, nc = non-conservative, ext=external forces


energy = friction
May be you meant work done by friction is considered as internal energy.
Well if friction is an internal force in the system you consider, work done by it will count under internal energy of the system. If it is an external force, work done by it will come under external energy of the system.

flyingpig

But - mg(h - h₀) is the work done by gravity

Abdul Quadeer

Yes it is.
Using the equation
Ui + Ki = Uf + Kf,
Ki - Kf = Uf - Ui
½m(v₀² - v²) = Uf - Ui

Now the change in potential energy of a system corresponding to a conservative internal force is defined as Uf - Ui = -W, where W is the work done on the system by the internal force (here gravity is the internal force if you consider earth as a part of your system) as the system passes from its initial configuration i to final configuration f.
So
½m(v₀² - v²) = -[- mg(h - h₀)]
½m(v² - v₀²) + mg(h - h₀) = 0
Again in accordance with the equation I gave.

flyingpig

All this internal and external system really confuses me...

Abdul Quadeer

To apply conservation of energy principle in questions, you have to select your system first. Here is an example-

A block of mass 4 kg starts falling with an initial velocity of 4m/s from a height of 10m. What is is velocity when it falls through a height of 2 m?

Take the block + the earth as the system. Only the block falls, so only the work done on the block will contribute to the gravtitational potential energy. As is descends through a height 8m, the potential energy decreases by mgh. No external force is acting on the system (here gravitational force is an internal force as we already took earth in our system).

So using conservation of energy,
increase in kinetic energy = decrease in potential energy
0.5m(v_f² - v_i²) = mg(h_i - h_f)

Which is same as

mg10 + 0.5m(4)² = mg2 + 0.5mv² (taking potential energy on the surface of earth as 0)
On solving, v=13.27m/s


If you still don't get the concept, try reading some good books like Resnick Halliday Walker Fundamentals of Physics, Sears and Zemansky University Physics etc to develop your fundamentals. Moreover, you can't understand physics unless you think about the subject.

PhanthomJay

A spring force is a conservative force. You can most of the time just use the general conservation of energy equation:

K_i +U_gi + U_si + W_nc = K_f +U_gf + U_sf

Where the subscripts are defined as i = initial, f = final, g = gravitational, s =spring, and where W_nc is the all encompassing work done by non-conservative external forces, like friction forces, applied forces, tension forces, normal forces, etc.

Doc Al

What confused you even more? My statement that gravitational PE increases when an object is raised?

For this problem, internal energy is irrelevant. The change in KE of the object equals the work done on the object by all forces. There are three forces acting on the object that do work: gravity, the applied force, and friction.

One key difference is that here you are treating gravity as a force, whereas before you represented gravity using gravitational PE. Either way is fine. A conservative force such as gravity can be represented as a potential energy.
Not sure what you are talking about. For this problem, all you care about is the kinetic energy of the object. (It's true that the work done by friction ends up as random thermal energy, but that's irrelevant.)
Not sure what you're getting at here. Depending upon how you define your system, you could consider spring PE as an internal energy. But in the same way that you can represent gravity either as a force or an energy, you can represent the action of the spring as either a force or an energy (since the spring force is conservative).

You have two choices in solving problems with conservative forces:

(1) Treat them as forces: ΔKE = Work done by all forces

(2) Treat them in terms of PE: ΔKE + ΔPE = Work done by non-conservative forces

The first is called the "Work-Energy" theorem; the second is called "conservation of energy". But they are equivalent (almost).

In order to use these equations intelligently, you must realize, for example, that the work done by gravity is -mgΔh and the change in gravitational PE is mgΔh. (The sign must change since the gravity term moves from one side of the equation to the other.)

flyingpig

Here is here I got confused, why is there another negative sign?

Shouldn't you leave it as it is and so?

-½m(v² - v₀²) = - mg(h - h₀)

So that

-ΔKE = ΔPE

But it doesn't make sense when you have

-ΔPE - ΔKE = 0

Since I said ΔPE = - mg(h - h₀)

- ( - mg(h - h₀) ) -½m(v² - v₀²) = 0

mg(h - h₀) - ½m(v² - v₀²) = 0