# Work, PE, and KE

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#### MedPR

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10+ Year Member
Can someone clear this up for me.

I understand that Work = Fdcostheta = delta KE = KEf - KE i
I understand that KEi + PEi = KEf + KEf

However, I'm not sure of the relationship between work and PE.

There is an example in NOVA that explains if you push a block with constant velocity from rest, up an incline, to rest, that the work done is 0. That's fine, constant velocity = 0 acceleration = 0 F = 0 Work.

However, the box gains potential energy, and don't you have to put in some form of energy to get the potential energy gained by the box? If KE initially was 0, and Work is 0, where did the PE come from?

Also, what is the relationship between Work and PE?

so you are saying that even though PE is increasing ... KE remains constant (even though it must decrease for law of conservation of energy). Well, since the object is not slowing down - constant velcotiy - KE is same because some force is being applied to keep that speed constant and that force is not zero. That force is being used to do work against friction etc. However, Fnet is 0 because u said conc vel = 0 acc etc

so you are saying that even though PE is increasing ... KE remains constant (even though it must decrease for law of conservation of energy). Well, since the object is not slowing down - constant velcotiy - KE is same because some force is being applied to keep that speed constant and that force is not zero. That force is being used to do work against friction etc. However, Fnet is 0 because u said conc vel = 0 acc etc

The example said from rest, constant velocity, finish at rest. It didn't say there was a gain in potential energy, but since it moved up the incline the mgh increased, so it must have gained PE.

Can someone clear this up for me.

I understand that Work = Fdcostheta = delta KE = KEf - KE i
I understand that KEi + PEi = KEf + KEf

However, I'm not sure of the relationship between work and PE.

There is an example in NOVA that explains if you push a block with constant velocity from rest, up an incline, to rest, that the work done is 0. That's fine, constant velocity = 0 acceleration = 0 F = 0 Work.

However, the box gains potential energy, and don't you have to put in some form of energy to get the potential energy gained by the box? If KE initially was 0, and Work is 0, where did the PE come from?

Also, what is the relationship between Work and PE?

i agree with you that there should be an increase in change in PE however do remember reading somewhere that
-(change in PE) = W

is it possible that PE is 0?

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i agree with you that there should be an increase in change in PE however do remember reading somewhere that
-(change in PE) = W

is it possible that PE is 0?

If h>0, then PE > 0 as far as I know.

W= -deltaPE? Hmm.. I don't think I've read that anywhere.

It makes sense though, right?

If W = KEf-KEi

and KEi + PEi = KEf + PEf

then KEf - KEf = PEi-PEf

and PEi - PEf = - deltaPE

I'll have to take a second look at the example I'm referring to.. Maybe I'm missing something. Will update soon.

Here it is.

A woman is pushing a cart of mass m slowly at a constant speed up an incline which makes an angle theta with the horizontal. The cart goes from the floor level to a height h.

Question a: How much work does the woman do in terms of m, g, h, and theta?
Question b: What is the total work done on the cart?

Solution A: The words "constant speed" and "straight path" imply that the cart's acceleration is zero, and the net force on it is zero, so we write:
Fnetx=0
From figure 9-2 we obtain -- Figure 2 is just a generic incline plane figure where Fw is the force up the incline.
Fnetx = Fw-mgsintheta

Combining these equations gives us Fw-mgsintheta=0, or Fw=mgsintheta

We can find deltaX by trigonometry. If we look at the large triangle in Figure 9-2, then we have

deltax = h/sintheta (just SOH, here)

Thus, the work done by the woman on the cart is W=mgsintheta(h/sintheta) or W = mgh.

Solution B: What is the total work done? Well, since the speed and direction are constant, the acceleration is zero, and Fnet=0, so Work=0.

How is this possible? They go through the trouble of deriving how Work = mgh, which means Work = Potential Energy, but then they say that the work = 0. The woman pushed the cart up an incline thereby increasing h, and thus creating/increasing Potential Energy. How can you have a non-zero height and zero PE?

Here it is.

A woman is pushing a cart of mass m slowly at a constant speed up an incline which makes an angle theta with the horizontal. The cart goes from the floor level to a height h.

Question a: How much work does the woman do in terms of m, g, h, and theta?
Question b: What is the total work done on the cart?

Solution A: The words "constant speed" and "straight path" imply that the cart's acceleration is zero, and the net force on it is zero, so we write:
Fnetx=0
From figure 9-2 we obtain -- Figure 2 is just a generic incline plane figure where Fw is the force up the incline.
Fnetx = Fw-mgsintheta

Combining these equations gives us Fw-mgsintheta=0, or Fw=mgsintheta

We can find deltaX by trigonometry. If we look at the large triangle in Figure 9-2, then we have

deltax = h/sintheta (just SOH, here)

Thus, the work done by the woman on the cart is W=mgsintheta(h/sintheta) or W = mgh.

Solution B: What is the total work done? Well, since the speed and direction are constant, the acceleration is zero, and Fnet=0, so Work=0.

How is this possible? They go through the trouble of deriving how Work = mgh, which means Work = Potential Energy, but then they say that the work = 0. The woman pushed the cart up an incline thereby increasing h, and thus creating/increasing Potential Energy. How can you have a non-zero height and zero PE?

hmmm thats weird that they'd ask for W in terms of m g h and theta when work is 0....i

solution B is weird because
delta KE = - (delta PE)

except here
delta KE = 0 and
(delta PE) = - (constant(hf) - (constant)hi) which is NOT 0

.....weird

I am not sure but I would say they are trying to teach two different scenarios.

In the first question, they asked you for the Work Done BY the women to climb the incline and since women just covered a vertical distance .. her work done would be mgh.

In the second question, they are asking for Work Done ON the CART BY the woman ... so here we look at the and see the force acting on it (force with which the women is pushing it) and since this force is providing constant velocity ... a = 0 ... so no NET work is done ON the cart. However, I would still think that if they asked us that what is the PE of the cart at the top ... it would still be m_cart g h. But again there is no work done by the woman to bring that cart from bottom to top.

its a weird question nonetheless

hmmm thats weird that they'd ask for W in terms of m g h and theta when work is 0....i

solution B is weird because
delta KE = - (delta PE)

except here
delta KE = 0 and
(delta PE) = - (constant(hf) - (constant)hi) which is NOT 0

.....weird

Yea.. After they said Work = mgh = 0, I kept reading thinking (and hoping) they would make some correlation with PE.. something like "Even though work is 0, there is PE because..." But they didn't.

I thought in part A they are trying to show, even though she used the ramp, the work= mgh, just as she could have not use the ramp and decided to lift it up instead..... she pushed it up the hypotenuse of the triangle (of course this is in the case of frictionless) but still the work is the same kind of an statement!!!

and second part, they say the acceleration of the cart is 0 (of course acceleration by her), even though there is still gravity as she is pushing up the ramp!? there has to be a work done against gravity to lift that **** up somehow.. so idk

Everybody, we all need to chill out on this one.

Question A: How much work does the woman do to the cart? mgh
Question B: What is the total amount of work done on the cart? 0

Question C: How much work did GRAVITY do to the cart? -mgh
remember, as far as gravity is concerned, the angle between F (down) and d (up the incline) is between 90 and 180 degrees, so costheta and therefore Work is a negative number here

problem solved!

Everybody, we all need to chill out on this one.

Question A: How much work does the woman do to the cart? mgh
Question B: What is the total amount of work done on the cart? 0

Question C: How much work did GRAVITY do to the cart? -mgh
remember, as far as gravity is concerned, the angle between F (down) and d (up the incline) is between 90 and 180 degrees, so costheta and therefore Work is a negative number here

problem solved!

I don't get it

Everybody, we all need to chill out on this one.

Question A: How much work does the woman do to the cart? mgh
Question B: What is the total amount of work done on the cart? 0

Question C: How much work did GRAVITY do to the cart? -mgh
remember, as far as gravity is concerned, the angle between F (down) and d (up the incline) is between 90 and 180 degrees, so costheta and therefore Work is a negative number here

problem solved!

so whatever work the woman does is cancelled out by gravity ... wow .. never thought like that ... thanks ...

I don't get it

I think he is saying for:

How much work gravity does on the cart?

so if the force of gravity is acting down on the cart and the cart is moving up the incline .. we have WD = Fd cos theta ... and theta here would be the angle between force of gravity and the direction of the cart moving (up the plane) which would be greater than 90 degree and less than 180 degree. Cosine of anything between them would be negative. so it would be -mgh.

So, women does a work of mgh on the cart & the gravity does a work of -mgh on the cart. So net work done is zero.

hope that helped a bit

I guess I understand how the work done by gravity cancels out the work done by the woman.. but how can there be no PE? What negates PE when there is a non-zero h?

I guess I understand how the work done by gravity cancels out the work done by the woman.. but how can there be no PE? What negates PE when there is a non-zero h?

We are not saying that there is no PE. I think there still is PE of -mgh on the cart at the top but the NET work done on the cart is zero because the WD by gravity cancels the WD by the women. So there is no net work done on the cart in the system as a whole. fk i am confused as well

We are not saying that there is no PE. I think there still is PE of -mgh on the cart at the top but the NET work done on the cart is zero because the WD by gravity cancels the WD by the women. So there is no net work done on the cart in the system as a whole. fk i am confused as well

I understood that the work was canceled out until NOVA defined work as w=mgh.

Again, you need to cafefully define your system. If your system is the woman, of course she did work. She applied a force (mg up) over a displacement (h up).

If your system is the block, it had no change in kinetic energy, so no NET work was done on the block. Well geez, the block had a force (the woman) applied over a distance (the ramp) so how come it didn't accelerate? F=ma and all that...

Remember, gravity applied a second force (mg down) over the same displacement (h up) so the net work, and therefore the change in kinetic energy of the block, is zero. Gravity did negative work on the block. This negative work is stored as gravitational potential energy.

NOVA is correct if they are calculating the work done by the woman. But not the (net) work done on the block.

Again, you need to cafefully define your system. If your system is the woman, of course she did work. She applied a force (mg up) over a displacement (h up).

If your system is the block, it had no change in kinetic energy, so no NET work was done on the block. Well geez, the block had a force (the woman) applied over a distance (the ramp) so how come it didn't accelerate? F=ma and all that...

Remember, gravity applied a second force (mg down) over the same displacement (h up) so the net work, and therefore the change in kinetic energy of the block, is zero. Gravity did negative work on the block. This negative work is stored as gravitational potential energy.

NOVA is correct if they are calculating the work done by the woman. But not the (net) work done on the block.

That clears everything up Thanks.