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I used delta K + delta U is nonconservative force.

They want kJ instead of J. :yuck:

I got -0.951375 kJ

but it tells me I'm wrong. I only have one more try. Could it be this number but positive?

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- Thread starter marcia888
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- #1

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I used delta K + delta U is nonconservative force.

They want kJ instead of J. :yuck:

I got -0.951375 kJ

but it tells me I'm wrong. I only have one more try. Could it be this number but positive?

- #2

- 429

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marcia888 said:

I used delta K + delta U is nonconservative force.

They want kJ instead of J. :yuck:

I got -0.951375 kJ

but it tells me I'm wrong. I only have one more try. Could it be this number but positive?

you may have made a mistake with the sign of "2.05m."

the change in potential energy is

mg h_final - mg h_initial

or, factoring...

mg (h_fin - h_in)

now, the surfer DROPS that given height. so his final position is LOWER than his initial position!

taking h_fin to be where h=0, you get...

delta PE = m*g*(-2.05m)

and, of course, that 8m/s speed is the final velocity, while the 1m/s speed is the initial.

this should give ya the right answer.

an intuitive way to get the sign correct is to think whether the external work added or took away from the systems energy. one way to determine this is to compare this situation to one where there is no external work.

(of course, this requires you to choose which new parameter you need to solve for. for instance, if you decide to keep the values given for inital velocity and height, the value of the final velocity will tell you whether the energy was diminished or increased. if the final velocity was lower without the external work, then the external work increased the energy and should be positive, etc.)

- #3

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Since the speed of the surfer is greater after the reaction, it is clear that his energy has increased. Some of the increase is due to gravity and some is due to the waves. It's certainly possible that the waves actually did negative work on him, they could have slowed him down a little bit during the reaction. So the answer is not neccesarily positive. But it turns out that the waves did actually do positive work.

m = mass

[tex]v_{1}[/tex] = initial speed

[tex]v_{2}[/tex] = final speed

h = height

g = acceleration due to gravity

The conservative work done is the gravitational work done, all other work is nonconservative, so:

Total Work Done - Gravitational Work Done = Nonconservative Work Done

To find the Total Work Done, you take the difference of the kinetic energies before and after.

Total Work Done:

[tex]= \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2}[/tex]

[tex]= \frac{1}{2}m(v_{2}^{2} - v_{1}^{2})[/tex]

Now from the Total Work Done, you subtract the Gravitational Work Done, which is equal to the Gravitational Potential Energy an object has when it is 2.05m off the ground.

Gravitational Potential Energy for 2.05m = Gravitational Work Done = Conservative Work Done = mgh

Total Work Done - Conservative Work Done = Total Nonconservative Work Done

[tex] = \frac{1}{2}m(v_{2}^{2} - v_{1}^{2}) - mgh[/tex]

[tex] = m(\frac{v_{2}^{2} - v_{1}^{2}}{2} - gh)[/tex]

I get a positive value > 10kJ

That help at all?

m = mass

[tex]v_{1}[/tex] = initial speed

[tex]v_{2}[/tex] = final speed

h = height

g = acceleration due to gravity

The conservative work done is the gravitational work done, all other work is nonconservative, so:

Total Work Done - Gravitational Work Done = Nonconservative Work Done

To find the Total Work Done, you take the difference of the kinetic energies before and after.

Total Work Done:

[tex]= \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2}[/tex]

[tex]= \frac{1}{2}m(v_{2}^{2} - v_{1}^{2})[/tex]

Now from the Total Work Done, you subtract the Gravitational Work Done, which is equal to the Gravitational Potential Energy an object has when it is 2.05m off the ground.

Gravitational Potential Energy for 2.05m = Gravitational Work Done = Conservative Work Done = mgh

Total Work Done - Conservative Work Done = Total Nonconservative Work Done

[tex] = \frac{1}{2}m(v_{2}^{2} - v_{1}^{2}) - mgh[/tex]

[tex] = m(\frac{v_{2}^{2} - v_{1}^{2}}{2} - gh)[/tex]

I get a positive value > 10kJ

That help at all?

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- #4

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subodei said:Since the speed of the surfer is greater after the reaction, it is clear that his energy has increased. Some of the increase is due to gravity and some is due to the waves. It's certainly possible that the waves actually did negative work on him, they could have slowed him down a little bit during the reaction. So the answer is not neccesarily positive. But it turns out that the waves did actually do positive work.

m = mass

[tex]v_{1}[/tex] = initial speed

[tex]v_{2}[/tex] = final speed

h = height

g = acceleration due to gravity

The conservative work done is the gravitational work done, all other work is nonconservative, so:

Total Work Done - Gravitational Work Done = Nonconservative Work Done

To find the Total Work Done, you take the difference of the kinetic energies before and after.

Total Work Done:

[tex]= \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2}[/tex]

[tex]= \frac{1}{2}m(v_{2}^{2} - v_{1}^{2})[/tex]

Now from the Total Work Done, you subtract the Gravitational Work Done, which is equal to the Gravitational Potential Energy an object has when it is 2.05m off the ground.

Gravitational Potential Energy for 2.05m = Gravitational Work Done = Conservative Work Done = mgh

Total Work Done - Conservative Work Done = Total Nonconservative Work Done

[tex] = \frac{1}{2}m(v_{2}^{2} - v_{1}^{2}) - mgh[/tex]

[tex] = m(\frac{v_{2}^{2} - v_{1}^{2}}{2} - gh)[/tex]

I get a positive value > 10kJ

That help at all?

the answer is positive, but it was the positive of the answer he had given!

the correct formula is, as he listed:

change in Kinetic Energy + change in Potential Energy = external (or nonconservative) Work

the textbook he probably uses is by serway and faughn, college physics; that's the precise equation they have in there for this!

- #5

Doc Al

Mentor

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That should give you the net work done on the surfer by nonconservative forces. Show what you got for [itex]\Delta K[/itex] and [itex]\Delta U[/itex]. (The change in K is positive, but the change in U is negative.)marcia888 said:I used delta K + delta U is nonconservative force.

Actually, yes. But how did you get a negative sign? (Remember that [itex]\Delta X = X_{final} - X_{initial}[/itex].)They want kJ instead of J. :yuck:

I got -0.951375 kJ

but it tells me I'm wrong. I only have one more try. Could it be this number but positive?

- #6

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Brad Barker said:the answer is positive, but it was the positive of the answer he had given!

the correct formula is, as he listed:

change in Kinetic Energy + change in Potential Energy = external (or nonconservative) Work

the textbook he probably uses is by serway and faughn, college physics; that's the precise equation they have in there for this!

The formula I used is the same thing.

Change in kinetic energy (KE after - KE before) plus the change in potential energy (-mgh)

I got 950kJ however, not .950 kJ

Can you see where I made an error?

(Btw on a different note, don't you think it's a little stupid to have a formula which really just says "Any change in energy from outside of the system is nonconservative"? Seems to me there is a better way to express it then have students memorize that [tex]\Delta{K} + \Delta{U}[/tex] = Nonconservative work.)

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- #7

Doc Al

Mentor

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Your formula is OK. Check your arithmetic.subodei said:The formula I used is the same thing.

Change in kinetic energy (KE after - KE before) plus the change in potential energy (-mgh)

I got 950kJ however, not .950 kJ

Can you see where I made an error?

- #8

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- #9

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V=V(q) for conservative & V=V(q,t) for nonconservative ?

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