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does work equal change in potential energy or change in kinetic energy?

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- Thread starter UrbanXrisis
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does work equal change in potential energy or change in kinetic energy?

- #2

arildno

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If you don't bother to make the distinction between conservative and non-conservative forces, then the work equals the change of kinetic energy.

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- #3

Doc Al

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I read that something =(-dPE)/(dt)

not I for got what. Any ideas? Power=(-dPE)/(dt)?

not I for got what. Any ideas? Power=(-dPE)/(dt)?

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UrbanXrisis said:I read that something =(-dPE)/(dt)

not I for got what. Any ideas? Power=(-dPE)/(dt)?

Power is [tex]\frac{work}{time}[/tex]

What you are reading is a calculus based derivative, not relavent to what you are asking.

- #6

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if you integrate it, it will become PE/t, which is work/time which is power.

What variable do you integrate to get power?

also, is this all correct?

∫F(x)=w(x)

∫P(t)=w(t)=Fv=F(at)=∆KE/t

- #7

arildno

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1) Let us consider Newton's 2. law for a system (particle if you like) with constant mass m under influence of a SINGLE force [tex]\vec{F}[/tex]:

[tex]\vec{F}=m\vec{a} (1)[/tex]

2) Let us take the dot product of (1) with the velocity [tex]\vec{v}[/tex]:

[tex]\vec{F}\cdot\vec{v}=m\vec{a}\cdot\vec{v} (2)[/tex]

Now, note that the right-hand side can be re-written as:

[tex]m\vec{a}\cdot\vec{v}=\frac{d}{dt}(\frac{1}{2}m\vec{v}^{2})=\frac{dK.E.}{dt}[/tex]

Or, introducing the POWER [tex]\mathcal{P}=\vec{F}\cdot\vec{v}[/tex]

[tex]\mathcal{P}=\frac{dK.E.}{dt} (3)[/tex]

(If we have a system under influence of several forces, (3) reads NET power equals rate of change of kinetic energy)

3) Let us integrate (3) between two arbitrary time values:

[tex]\int_{0}^{T}\vec{F}\cdot\vec{v}dt=K.E\mid_{t=T}-K.E\mid_{t=0}(4)[/tex]

But, since [tex]\vec{v}=\frac{d\vec{x}}{dt}[/tex]

we may rewrite the time integral with the line integral along the object's trajectory I between positions [tex]\vec{x}(0),\vec{x}(T)[/tex]

That is:

[tex]\int_{0}^{T}\vec{F}\cdot\vec{v}dt=\oint_{I}\vec{F}\cdot{d\vec{x}}=W (5)[/tex]

That is, we have:

[tex]W=\bigtriangleup{K.E}(6)[/tex]

(where W is the work done)

(A notation which ought to be clear)

And also:

[tex]\frac{dW}{dt}=\mathcal{P}(7)[/tex]

Clear so far?

- #8

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okay, so

∫F(x)=w(x)

∫P(t)=w(t)

P=Fv=F(at)=∆KE/t

W= ∆KE

does ∫W=-dPE/dt?

∫F(x)=w(x)

∫P(t)=w(t)

P=Fv=F(at)=∆KE/t

W= ∆KE

does ∫W=-dPE/dt?

- #9

arildno

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Do you know the difference between conservative and non-conservative forces?

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[tex]\int\mathcal{F}(x)=w(x)[/tex]

where x is in terms of meters

[tex]\int\mathcal{P}(t)=w(t)[/tex]

where t is in terms of time

arildno said:

Do you know the difference between conservative and non-conservative forces?

energy is lost in motion?

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- #12

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this is:

Power=Force*velocity

Power=force*(acceleration*time)

Power=work/time

Power=change in kinetic energy/time

- #13

arildno

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No, and your notation is still not good.

A conservative force has the following property:

We say that a force is CONSERVATIVE if the work done by that force ONLY DEPEND UPON THE END POINTS OF THE TRAJECTORY, and NOT on the particular trajectory connecting those two points.

That is, suppose your "initial point" is [tex]\vec{x}_{0}[/tex] and your final point is [tex]\vec{x}_{T}[/tex] (that is, just two points in space)

Now, let I and J be two different trajectories between [tex]\vec{x}_{0},\vec{x}_{T}[/tex]

For a conservative force it is then true that:

[tex]\oint_{I}\vec{F}\cdot{d\vec{x}}=\oint_{J}\vec{F}\cdot{d\vec{x}}[/tex]

for ANY choice of connecting trajectories I, J.

That is, the value of the work done can ONLY depend on [tex]\vec{x}_{0}[/tex] and [tex]\vec{x}_{T}[/tex]

Okay with that?

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so what are some examples of non conservative forces vs conservative forces?

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arildno

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Take for example the force of gravity (a CONSERVATIVE force):

[tex]\vec{F}=-mg\vec{k}[/tex]

Now,

[tex]\vec{F}\cdot\vec{v}=-mg\frac{dz}{dt}[/tex]

(where z(t) is the vertical position as a function of time)

Hence, the work done by gravity can be written as:

[tex]W=\int_{0}^{T}-mg\frac{dz}{dt}dt=-mgz(T)+mgz(0)=-mgz_{T}+mgz_{0}[/tex]

That is, IRRESPECTIVE of the path the object has undertaken, the work done by gravity is simply found as the difference between final vertical position and initial vertical position multiplied with (-mg)

AGREED?

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- #17

arildno

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It means that the value of the WORK doesn't depend upon the particular PATH you may have gone along, but ONLY on the initial and final POSITIONS.

In particular, it implies that if you end up where you started NO NET WORK has been produced by that force!

To see this, I just state that we mathematically can prove, that a conservative force

can ALWAYS be regarded as the gradient of a scalar potential -U, or, in the 1-D case:

[tex]F=-\frac{dU}{dx}[/tex] (in general, [tex]\vec{F}=-\nabla{U}[/tex]

Let's take the case of gravity:

We have now found, by equating with the change in kinetic energy:

[tex]-mgz_{T}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(T)=\vec{x}_{T})}- \frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(0)=\vec{x}_{0})[/tex]

Rearranging, we have:

[tex]\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{0}}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{T}}+mgz_{T}[/tex]

Since 0, T are ARBITRARY, doesn't this merely state CONSERVATION OF MECHANICAL ENERGY?

(Since only gravity were assumed to be present, the velocities in the horizontal directions haven't changed, so they may be removed from your equation)

In particular, it implies that if you end up where you started NO NET WORK has been produced by that force!

To see this, I just state that we mathematically can prove, that a conservative force

can ALWAYS be regarded as the gradient of a scalar potential -U, or, in the 1-D case:

[tex]F=-\frac{dU}{dx}[/tex] (in general, [tex]\vec{F}=-\nabla{U}[/tex]

Let's take the case of gravity:

We have now found, by equating with the change in kinetic energy:

[tex]-mgz_{T}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(T)=\vec{x}_{T})}- \frac{1}{2}m\vec{v}^{2}\mid_{(\vec{x}=\vec{x}(0)=\vec{x}_{0})[/tex]

Rearranging, we have:

[tex]\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{0}}+mgz_{0}=\frac{1}{2}m\vec{v}^{2}\mid_{\vec{x}_{T}}+mgz_{T}[/tex]

Since 0, T are ARBITRARY, doesn't this merely state CONSERVATION OF MECHANICAL ENERGY?

(Since only gravity were assumed to be present, the velocities in the horizontal directions haven't changed, so they may be removed from your equation)

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what is a nonconservative force then?

When does it depend on the path?

When does it depend on the path?

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- #19

arildno

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Air resistance is one example (that is also a DISSIPATIVE force, in that it removes mechanical energy from the system (or if, you like, the object loses momentum to the air molecules which get set in motion))

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If a ball was allowed to slide down an inclice, then the force of friction is a non-conservative force?

is the net work = 0?

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arildno

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I'll take the friction force on the incline:

The general rule is ALL dissipative forces should be regarded as non-conservative forces.

This is because they always EXTRACT energy from your system, NEVER impart kinetic energy.

You will have verified that a force is NON-conservative, if you can choose two points, connect them with two different paths, and demonstrate that the work done by the force differ on the two paths.

To do this, let us set up the following situation:

First, we let the ball slide a distance D1 along the incline, stop the ball with our hand, and then push it back up the incline to the initial position (we push it just tangentially, so that we don't increase the normal force or, hence the friction force).

The friction force always opposes the direction of motion, so the net work it has done on this path is typically [tex]\mu{N}2D_{1}[/tex]

We repeat the experiment with a different distant D2; the total work of the friction force is on this path: [tex]\mu{N}2D_{2}\neq\mu{N}2D_{1}[/tex]

The path [tex]0\to{D_{1}}\to{0}[/tex] is a different trajectory than [tex]0\to{D}_{2}\to{0}[/tex]

That is, the work done by the friction depends on THE PATH LENGTHS, and NOT on the initial and final positions.

This is COMPLETELY different from the work done by the force of gravity.

For both scenarios, the work from the gravity force is:

[tex]W=-mgz(T)+mgz(0)=0[/tex]

Since z(T)=z(0) (we end up in the same height as were we started out)

That is, the work from gravity is ZERO on both paths

The general rule is ALL dissipative forces should be regarded as non-conservative forces.

This is because they always EXTRACT energy from your system, NEVER impart kinetic energy.

You will have verified that a force is NON-conservative, if you can choose two points, connect them with two different paths, and demonstrate that the work done by the force differ on the two paths.

To do this, let us set up the following situation:

First, we let the ball slide a distance D1 along the incline, stop the ball with our hand, and then push it back up the incline to the initial position (we push it just tangentially, so that we don't increase the normal force or, hence the friction force).

The friction force always opposes the direction of motion, so the net work it has done on this path is typically [tex]\mu{N}2D_{1}[/tex]

We repeat the experiment with a different distant D2; the total work of the friction force is on this path: [tex]\mu{N}2D_{2}\neq\mu{N}2D_{1}[/tex]

The path [tex]0\to{D_{1}}\to{0}[/tex] is a different trajectory than [tex]0\to{D}_{2}\to{0}[/tex]

That is, the work done by the friction depends on THE PATH LENGTHS, and NOT on the initial and final positions.

This is COMPLETELY different from the work done by the force of gravity.

For both scenarios, the work from the gravity force is:

[tex]W=-mgz(T)+mgz(0)=0[/tex]

Since z(T)=z(0) (we end up in the same height as were we started out)

That is, the work from gravity is ZERO on both paths

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- #22

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for energy to be conserved, you need KinEn1+PotEn1=KinEn2+PotEn2 right?

when the ball slides down the incline, it loses KinEn due to the force of friction. So this is why we call the Force if Friction a nonconservative force. right?

- #23

arildno

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That is the equation for conservation of mechanical energy, right.

The work done by a CONSERVATIVE force (for example, gravity) can always be written as the difference between potential energy levels (in the case of gravity, the difference between the values of the gravitational potential).

"when the ball slides down the incline, it loses KinEn due to the force of friction. So this is why we call the Force if Friction a nonconservative force. right?"

Not strictly correct.

If you throw a ball vertically upwards, the force of gravity "sucks" kinetic energy out it, or if you like, CONVERTS that kinetic energy into potential energy.

On the way down, the force of gravity IMPARTS kinetic energy to the ball, or if you like, CONVERTS potential energy into kinetic energy.

The force of friction always works opposite the direction of motion.

This means that the POWER from the force of friction can be written as:

[tex]\mathcal{P}=-||\vec{F}||||\vec{v}||[/tex]

where the magnitudes of the friction force and velocity is numbers greater or equal to zero.

But this means, since power is the time-derivative of work of the force:

[tex]\frac{dW}{dt}<0[/tex]

Considering now a situation in which the friction force is the only force, the power/rate of change of kinetic energy-equation says:

[tex]\frac{dW}{dt}=\frac{dK.E}{dt}<0[/tex]

That is, the friction force is necessarily DISSIPATIVE, always reducing the kinetic energy of the system.

Since every dissipative force is non-conservative, the force of friction is non-conservative.

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