# Work, Energy, Power, Momentum.

#### PrudensOptimus

Can someone distribute their knowledge of the above topics? I have a test on these tomorrow.

Thanks.

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#### turin

Homework Helper
WORK: Work is the transfer of mechanical energy
ENERGY: Energy is the potential to do work. It is conserved when you account for all of its forms, in a closed inertial system.
POWER: Power is the time rate at which work is done.
MOMENTUM: Momentum is the quantity of motion (or ability to provide impulse). It is conserved both linearly and rotationally in a closed system in an inertial frame.

WORK can be calculated in two basic ways:

1) Integrate the dot product of the force with the position displacement over the path. This has a simplification that comes up in first semester physics: W = Fd, where W is the work, F is the force in the direction of travel (assumed to be constant, and here, basically treated as a scalar, not a vector), d is the distance traveled. Note that W can be negative (the object had to do work in order to move) if the force is in the oposite direction of the travel.

2) Calculate the change in mechanical energy. Usually this will be simpler, so, if you are given a bunch of energy info, then you probably want to use this one. W = DeltaKE + DeltaU = KE_final - KE_initial + U_final - U_initial, where W is the work (done to the object), KE is the kinetic energy (1/2 m v^2), U is the potential energy (mgh for gravity, 1/2 k x^2 for a spring). Note that W can be negative for the same reason as before (it could also be positive). This indicates that the system is not conservative (usually this means that friction is present). If the work is zero, then the system is either conservative, or work is being done to the object to compensate for the work that it is doing. I would expect you to encounter the former case rather than the later.

Energy will probably just be the sum of the kinetic energies and potential energies. There are tons of details here. Basically, kinetic energy is translational and rotational (1/2 m v^2 and 1/2 I w^2). Potential will probably be gravitational or elastic (mgh or 1/2 k x^2). Remeber that the h is arbitrary for gravitational, but the x IS NOT ARBITRARY FOR THE ELASTIC POTENTIAL. IT IS THE DISTANCE FROM THE EQUILIBRIUM POSITION. It provides a resotring force, whereas gravity does not (usually, unless you're talking about a pendulum). Like i said, there are a lot of details for energy.

To calculate power, divide the amount of work done by the time it took to do the work. Also, there is thrust, which is the force of the exhaust gas multiplied by the velocity that it is flying out of the jet engine. I feel like I'm missing something important here, but I'll move on.

The rate of change of momentum of an object is defined to be the net (or resultant) force on an object (a generalization of Newton's second law). The change in momentum of an object is defined to be the impulse on an object. Momentum, force, and impulse are all VECTOR QUANTITIES. It is easy to screw up on a calculation when you use the magnitudes of these quantities without considering their direction.

Impluse can be calculated in two ways (notice the analogy to work):

1) It is the integral of the force (as a function of time) over time. This sometimes simplifies to I = Ft, where I is the impulse, F is the force (assumed to be constant), t is the duration of the force. Note that, since F is a vector, impulse will be a vector. (For work, F loses all of its vector like qualities except one: it can cause a negative result.)

2) Calculate the change in momentum: I = p_final - p_initial. Usually, it is safe to say that p = mv (for nonrelativistic particles, so unless otherwise stated in the problem, use this formula).

energy is to work is to power (not vectors, scalars)
as
momentum is to impulse is to force (vectors)

Work is what breaks the conservation of energy.

Impulse is what breaks the conservation of momentum.

That is not to say that whenever you have work or impulse, that energy and momentum are not conserved. If you consider work or impulse inside the system, then energy and momentum are conserved. If you consider work and energy as acting on the system from outside, then energy and momentum can break conservation in the system.

Example for energy:
If we consider a mass falling due to gravity, then we can say that gravity is doing work on the object because it is applying a downward force to the object, and the object is moving downward. The force is mg, the distance it travels is y, therefore the work done on the object is mgy (Fd). This is manifest in the fact that the object is speeding up as it falls (and thus gaining kinetic energy). Now, consider the same situation, but include the gravitational potential in the system. In this view, the increase in kinetic energy is due to the decrease in gravitational potential energy. Thus, the energy is conserved. In the first point of view, we treated gravity as external to the system. In the second point of view, we included the potential energy of gravity as part of the system.

Example for momentum:
This one is a bit more tricky. I will use a bit more obscure example, because I can't think of a more reasonable one. A ball hits a wall and bounces back. Its velocity has changed, and therefore its momentum has changed. Does this violate the conservation of momentum. No. Actually, the wall recoiled ever so slightly, but, since it is rigidly connected to the earth, and the earth is so huge, then the recoil was negligible. So, the change in mometum of the ball was mV, and the change in the momentum of the earth was Mv, and mV must equal Mv, but, since M >> m, then v << V.

Another example for momentum would be the falling object again. Its downward momentum increases because the force of gravity puts an impulse on it of Ft = mgt. Thus, it seems like the impulse breaks the conservation of momentum. But actually, the earth's momentum is again the compensation. The earth moves upward to meet the object as it falls. The impulse on the earth is -mgt. Since M >> m, then, again, the change in velocity isn't noticible.

#### turin

Homework Helper
I forgot to mention rotational stuff:

angular moment of inertia and angular velocity is to angular momentum is to torque
as
mass and velocity is to momentum is to force

(Usually, when you just hear "momentum" and "velocity" without a pre-qualifier, that refers to the linear version)

Symbolically:

Iw : L : &tau; :: mv : p : F

Angular momentum is conserved, separately from linear momentum (in the absence of a torque applied from outside the system).

Don't forget your centripetal acceleration equations:

a_c = m (v^2) / r = m (w^2) r

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#### PrudensOptimus

Wow, looks yummy, thanks for your share of knowledge, Turin.

I tried to integrate Work, &int;&Sigma;F dx, where x is displacement, replacing F with ma, I end up getting &delta;KE&int;1/dx dx. Does &int;1/dx dx give you ln|x| + C, or something else? I'm not sure. And what is the bottom and top index of the Work Integral? Is it the initial position and the final posistion?

#### PrudensOptimus

When do you use Wnet = &Delta;KE?

I saw some questions where W = &Delta;KE + &Delta;PE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = &Delta;KE, the W must mean for conservative?

#### PrudensOptimus

What does it mean:

"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term WNC, on the left(but not in both!)"

EQ 6-10 : WNC = &Delta;KE + &Delta; PE.

#### turin

Homework Helper
Originally posted by PrudensOptimus
I tried to integrate Work, &int;&Sigma;F dx, where x is displacement, replacing F with ma, I end up getting &delta;KE&int;1/dx dx.
I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.

Is this what you were trying to do?

&int;Fdx = &int;madx = m&int;(dv/dt)dx = m&int;(dv/dx)(dx/dt)dx
= m&int;(dv/dx)vdx = m&int;vdv = (m/2)[v2 - v02] = KE - KE0 = &Delta;KE

#### turin

Homework Helper
Originally posted by PrudensOptimus
When do you use Wnet = &Delta;KE?

I saw some questions where W = &Delta;KE + &Delta;PE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = &Delta;KE, the W must mean for conservative?
You probably use this notion when you are considering the work being done by the potential energy in the problem. If you include the potential energy in the problem, then the system is conservative. If you define the system without the potential energy (i.e. call gravity an external force), then the system is not conservative.

Also, for instance, when you have an object sliding along some surface, and there is friction, then the friction will slow it down. Since the object slows down, it will lose kinetic energy, but there really isn't any potential energy to account for (the energy changes into thermal energy, the mechanism is called heat). So, friction usually makes the system non-conservative. To calculate the work done by friction, it is usually easiest to use the equation that you have up there. You would use the first equation if there was no change in any potential energy. If, say, you have a block sliding down an inclined plane, you would need to use the second equation, because the gravitational potential energy is part of the mechanical energy, and thus, it will increase the kinetic energy.

#### turin

Homework Helper
Originally posted by PrudensOptimus
What does it mean:

"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term WNC, on the left(but not in both!)"

EQ 6-10 : WNC = &Delta;KE + &Delta; PE.
It looks like they are using the subscript, "NC," to indicate that they are strictly talking about "Non-Conservative" work. That means that there is no potential energy that you can associate with it. Friction, I'm pretty sure, will contribute to the WNC. Gravity, and spring force stuff, goes into the &Delta;PE on the right.

It's a little tricky, though, because it depends on how you define your system. I would say that it really doesn't matter, as long as you stick with the same defined system through your calculations.

The first law of thermodynamics says that &Delta;E = Q - W, where E is the energy in the system (accounting for ALL forms), Q is the heat transfered into the system, and W is the work done by the system. Maybe you will see the first law written as &Delta;E = W - Q. In this case, you just change the directions of W and Q. The sign just tells you which way the work and heat is going (into the system or out of the system). If you've seen this thermo stuff, then that WNC almost definitely means the W or -W in the first law of thermodynamics, assuming Q = 0, and all the energy is considered as kinetic and potential energy.

Sorry, I said "second" law when I meant "first" law. I have editted the mistake.
I changed some other stupid statements as well. I hope I didn't screw you up on your test.
Oh hell, I don't really know what to say about this. You would probably do best to ignore the thermodynamics part; I'm confusing myself with it.

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#### PrudensOptimus

Originally posted by turin
I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.

Is this what you were trying to do?

&int;Fdx = &int;madx = m&int;(dv/dt)dx = m&int;(dv/dx)(dx/dt)dx
= m&int;(dv/dx)vdx = m&int;vdv = (m/2)[v2 - v02] = KE - KE0 = &Delta;KE

When I replaced F with ma, and a with (v^2 - v0^2)/2&Delta;x, isn't m((v^2 - v0^2)/2) suppose to be treated as constants? and thus getting &Delta;KE * &int;1/&Delta;x dx?

#### PrudensOptimus

mç(dv/dx)vdx --- how did you get to that part?

#### turin

Homework Helper
Originally posted by PrudensOptimus
When I replaced F with ma, and a with (v^2 - v0^2)/2&Delta;x, isn't m((v^2 - v0^2)/2) suppose to be treated as constants?
F is the integrand, and it can be a function of the position of the particle, in general. If you replace it by ma, then the a must be a function of the position of the particle, in general. You cannot replace this by a constant, in general.

#### turin

Homework Helper
Originally posted by PrudensOptimus
mç(dv/dx)vdx --- how did you get to that part?
This is the chain rule in action:

Velocity, at a particular instant of time, has a value. You can make a convenient rule for this value which will accept, as its input, the time, and will give, as its output, the value of the velocity. This rule is the time derivative of the position of the particle, evaluated at the specific time input into the rule. In other words, this is the function, f(t):

v = f(t)

Velocity, at a particular point in space, has a value (because we're talking about a trajectory). I couldn't tell you off the top of my head what exactly that rule is, or where it comes from, without mentioning energy, but lets see what happens when we differentiate this rule (the function, g(x)) with respect to time:

v = g(x)

Now for the chain rule:

d[g(x)]/dt = d[g(x(t))]/dt

Here, I have just explicated the time dependence. Now, using the chain rule:

= g'(x(t))(dx/dt) = (dg/dx)(dx/dt)

But, going back to the first rule for assigning a value to v, you can see that:

(dx/dt) = v

so that:

d[g(x)]/dt = (dg/dx)v

But, g(x) also equals v at this point on the trajectory, so:

d[v]/dt = (dv/dx)v

But,

dv/dt = a

Thus,

a = v(dv/dx).

This is usually done a bit sloppily, referring to the velocity itself as an inherrent function of position and time. I think that doing so causes confusion. It is just the chain rule, but try to remember that we're talking about functions that return a value that is v, not v as a function in itself. After having said this, my notation is probably a bit sloppy. To try to make the point a bit more clear:

We usually talk about v as a function of t, and we denote it:

v = v(t)

This is the f(t).

What we want to end up with is an expression with a, v, and x, without t hanging around and bothering us. What we do in terms of the math is to start with an integral of a function of t over the variable x. This is messy. We could express x in terms of t, but we choose t in terms of x. We use the chain rule to get the time derivative of f(t) in terms of a function, g(x), and its derivative. So, I should probably give it to you like this:

h(x) = g(x)(dg/dx)

where, h(x) gives the acceleration at the point, x:

a = h(x)

and g(x) gives the velocity at the point x

v = g(x).

So:

df/dt = g(x)(dg/dx).

It should be emphasised that this does not generally happen with the chain rule that the function just kind of pops out of the derivative. This occurs in this case because of the relationship between the two choices of independent variable that we're playing around with to each other.

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