Conserved Energy in a moving frame of reference

[Tinhorn]

Homework Statement

When the box is at the bottom of the incline it will have a velocity of v_f
The person is an inertial frame of reference with a velocity of v_f

When the person starts moving, the box is moving backwards with the velocity of v_f.

This means at the top of the hill the box has both potiental energy and kinetic energy
from the person's point of view.

But when it get to the bottom of the hill, the box has lost all its potiental energy.
It has also lost its kinetic energy because the box and the person are moving at the same velocity.

so now$\frac{mv^{2}}{2}$ + mgh = 0 at the top of the hill

At the top of the hill it had both Kinetic and Potiental energy and at the bottom, it doesn't have any energy.

Homework Equations

I'm basically supposed to prove that mgh = $\frac{mv^{2}}{2}$ is still true for an inertial frame of reference

The Attempt at a Solution

At a stationary frame of reference, normal work does no work.
mgcos($\Theta$). Theta always being 90 degrees

but in a moving frame of reference, the displacement is not the same. because to the person
the box looks like its moving left, slowing down and finally catching up. So the angle between displacement and Normal force is no longer 90 degrees.
So in this case Normal Force is definitely doing work.

I'm just stuck in proving that the work normal force does can now make up for the apparent loss of energy.
Meaning:
Finding the work normal force does make $\frac{mv^{2}}{2}$ + mgh = 0 into mgh = $\frac{mv^{2}}{2}$

What equations should i be looking at ,
Can you point me in the right direction

If you need clarification on something please tell me

 

[Tinhorn]

I hope I can bump this with additional work i did.

So i used the Conservation of Mechanical Energy

$KE_{i}$ + $PE_{i}$ + External forces = $KE_{f}$ + $PE_{f}$

Since Normal force is doing work. i put it down and solved for Normal Force

$\frac{mv^{2}}{2}$ + mgh + mgcosθ = 0

{in the final equation KE + PE = 0, we are trying to prove otherwise}

$\frac{mv^{2}}{2}$ + mgcosθ = - mgh

$mv^{2}$ + mgcosθ = - 2mgh


$v^{2}$ + mgcosθ = - 2gh

v + mgcosθ = $\sqrt{- 2gh}$

Law of conservation of energy: $\frac{mv^{2}}{2}$ = mgh
solve for v
v = $\sqrt{- 2gh}$

so Normal force = -2 $\sqrt{- 2gh}$

I don't know where to go from there

 

[PhanthomJay]

There are several errors you are making. Your energy equation is wrong , you noted that the initial KE + PE + external force = final KE + PE. The term "external force" should be instead "work done by non conservative forces" (like work done by the normal force). But I don't see how the normal force could do work regardless of the reference frame, since it would have to generate heat, which it does not. You also note that there is no relative speed difference at the bottom of the slope, but this is Not true. The relative velocities are different, since the directions of each are not the same, so you have to look at the vector difference, and the relative speed is the magnitude of the relative velocity. You might want to look first at the simpler case of a block dropped from rest while you are moving horizontally at Vf.

 

[Tinhorn]

Sadly i can't look at any simple cases
but The normal force is not perpendicular to the displacement anymore, that's why it is doing work

And they are are going in the same direction but of them are going to the right

 

[PhanthomJay]

Sadly i can't look at any simple cases
but The normal force is not perpendicular to the displacement anymore, that's why it is doing work

And they are are going in the same direction but of them are going to the right

I will look at the 'simple' case of a block of mass m initially at rest (with respect to a stationary observer) being dropped from rest to the ground from a height h above ground. We know that energy is conserved, so

$KE_i + PE_i = KE_f + PE_f$ , or
$0 + mgh = 1/2mv_f^2 + 0$
$mgh = 1/2mv_f^2$
$v_f = \sqrt{2gh}$

Now let's see whether energy is still conserved in this case when the observer is in an inertial reference frame moving horizontally at a constant speedv $v_f$ with respect to the ground. In this case, the initial speed of the block relative to the observer is $v_f$ to the left, as you have noted. Now when the block just reaches the ground, the block is moving at velocity $v_f$ vertically downward with respect to the ground, and the observer is moving at a velocity $v_f$ horizontally to the right with respect to the ground. So the relative velocity of the box with respect to the observer is the vector difference between these two velocities which are at right angle to each other. Thus, the vector difference per the rules for vector subtraction and using Pythagoras, is
$V_{bo} = \sqrt{v_o^2 + v_b^2}$
$V_{bo} = \sqrt{2v_f^2}$, (at an angle 45 degrees down and to the left).

Now applying conservation of energy

$1/2mv_f^2 + mgh = 1/2m(2v_f^2)$
$1/2mv_f^2 + mgh = mv_f^2$
$mgh = mv_f^2 - 1/2mv_f^2$
$mgh = 1/2mv_f^2$

Same result.

Which was to be proven.

Now extend this to your case of the block on an incline.

 

[Tinhorn]

Wouldn't be $V_{bo}$ be 0;
because initially the box is moving to the left because it just started moving but eventually when it gains speed. It will reach its maximum velocity $v_{f}$. The frame of reference is also moving at $v_{f}$. So the box looks like it is at rest.

 

[Tinhorn]

Do you mean

At the top for a stationary frame of reference
Energy = PE = mgh
but for moving reference
Energy = K.E + P.E
K.E = P.E because P.E turns into K.E

Energy = mgh + mgh
Energy = 2mgh

And can you give me hints on how much work normal force is doing exactly

 

[PhanthomJay]

Wouldn't be $V_{bo}$ be 0;
because initially the box is moving to the left because it just started moving but eventually when it gains speed. It will reach its maximum velocity $v_{f}$. The frame of reference is also moving at $v_{f}$. So the box looks like it is at rest.

No, this is not right. Now mind you we are talking about the vertical drop case, not the incline case. For one thing, if v{bo} was 0 when the block was at the bottom of its fall, then energy would not be conserved, but I can assure you that it definitely is. Rather, then, realize that although the block and observer are moving at the same speed at the bottom of the drop, relative to the ground, they are nor moving at the same speed relative to each other. If I'm moving at 5 m/s to the right wrt th ground and you're moving at 5 m/s to the left wrt the ground, then you are moving at 10 m/s to the left relative to me. If you are moving 5 m/s down wrt the ground and I'm moving 5 m/s to the right wrt the ground, then you are moving 5(sq root 2) or 7 m/s relative to me, at an angle 45 sdegrees down and to the left. Thus the relative speed is 7 m/s. It is not 0.

 

[Tinhorn]

Oh i get that for the vertical drop case

I was trying to apply it to the inclined case

 

[Doc Al]

Have you considered the forces that the block and incline exert on each other? Viewed from the moving frame, those forces do 'work'. Consider the displacement of the incline and the box in the moving frame.

 

[Tinhorn]

Have you considered the forces that the block and incline exert on each other? Viewed from the moving frame, those forces do 'work'. Consider the displacement of the incline and the box in the moving frame.

Yes the normal force.
I know the normal force does work because the angle between the displacement and the force is not 90 viewed from the moving frame
so mgcosθ is not 0
but i don't know exactly how much it does so that

mgh = $mv^{2}$/2

 

[Doc Al]

Try this: In the time it takes for the box to slide down the incline, what is the displacement of the incline and the box as seen by the moving frame? Then find the work done on each by the normal force.

 

[Tinhorn]

it kinda looks likes a half circle
First the box would be moving back and then slowing down and then acceleration

 

[Doc Al]

I don't quite understand your diagram.

In the rest frame, the box takes a diagonal path down the incline. In the moving frame, the box takes a different path, but it still goes from the top of the incline to the bottom.

 

[Tinhorn]

Does it move backwards first and then from the top of the incline to the bottom.

 

[Doc Al]

Does it move backwards first and then from the top of the incline to the bottom.

No, the box is always moving to the left, from the top of the incline to the bottom.

In looking over your first post, you are taking the speed of the moving frame to be equal to the speed of the box after it has left the incline and is moving horizontally. (That's how you get the speed of the box to be zero in the moving frame.)

It may be difficult to do the calculation of the work done, but here's the basic idea. The box and incline exert a normal force on each other, which ends up doing work. It does positive work on the incline, increasing its energy, and negative work on the box. The easiest way to figure out where the energy is going is not to worry about the details of the work done, but to use conservation of momentum. Then you can switch from one frame to the other and calculate the KE of each. You'll see that the incline gains just enough energy (in the moving frame) to account for the 'missing' energy and preserve conservation of energy.

 

[Tinhorn]

Lets say the velocity of the person is $V_{fr}$

and the velocity of the box is $V_{b}$


is the work on the incline m * $V_{fr}$ - $V_{b}$

and the work on the box the negative of that

 

[Tinhorn]

is the work done by both the negative force times the position

 

[Tinhorn]

isn't the force on the incline gravity

 

[Doc Al]

isn't the force on the incline gravity

No. The forces of interest are the normal force and the gravity that acts on the box.

View things from the 'rest frame', the frame in which the box is at rest on the incline just as it is released. Once the box slides down the incline, its speed is v. Use conservation of momentum to figure the speed of the incline/earth in that frame. Then you can transform to the moving frame. Rather than worry about the details of the normal force and work done, consider the KE in the two frames.

The key point is that in the moving frame, the incline/earth gains just enough energy to account for the 'missing' energy.

 

[Tinhorn]

So in the rest frame, the incline isn't moving.
So its KE = 0;

in the moving frame, the incline is moving with v.
so its KE = m$v^{2}$/2

 

[Doc Al]

So in the rest frame, the incline isn't moving.
So its KE = 0;

Actually, that's not quite true. Momentum conservation implies that the incline must move (at some small speed) opposite to the box. In the rest frame that additional KE is trivial, but it becomes significant in the moving frame.

 

[Tinhorn]

Do you know why the resultant force action on the box is this

$f_{r}$ = mgsinθcosθ x+ mg$sin^{2}$y

I thought only mgsinθ and mgcosθ
where did extra sin come in from

 

[Doc Al]

Do you know why the resultant force action on the box is this

$f_{r}$ = mgsinθcosθ x+ mg$sin^{2}$y

I thought only mgsinθ and mgcosθ
where did extra sin come in from

In the x direction, you have the x-component of the normal force.
In the y direction, you have both the y-component of the normal force and the weight.

 

[Tinhorn]

Can you explain that to me again
in greater detail please?

 

[Doc Al]

Can you explain that to me again
in greater detail please?

Only two forces act on the box: gravity and the normal force. Tell me the x and y components of each.

 

[Steely Dan]

To clarify what is being said here, the axes of the stationary and moving reference frames are still parallel to each other, and so the normal force is always perpendicular to the perceived motion. So this is equivalent to a situation where you move the incline to the left a certain number of meters; in that situation, the normal force is still perpendicular to the displacement moved and is not responsible for changing the energy of the box. It is true according to the mathematical definition of work if you define displacements with respect to the origin that the work done is non-zero, but it is obvious you can get the right answer by just shifting your reference frame at any given time.

 

[Tinhorn]

gravity has no x component
it just has mg sin theta

and normal force has
mg sin theta cos theta + mg cos ^2 theta

 

[Doc Al]

gravity has no x component
it just has mg sin theta

Why the sinθ? And what's the direction?

and normal force has
mg sin theta cos theta + mg cos ^2 theta

Good, but keep the x and y components separate. And mind the direction of each component.

To find the components of the net force, add the x components and add the y components.

 

[Tinhorn]

mg in the -y direction
that is why the sin is there

mg sin theta cos theta in the y
mg cos ^2 theta in the x

 

[Doc Al]

mg in the -y direction
that is why the sin is there

The weight acts downward, so is already totally in the y direction. No need for any sinθ.

mg sin theta cos theta in the y
mg cos ^2 theta in the x

I think you have these reversed.

 

[Tinhorn]

so mg for weight

and mg sin theta cos theta in the x
mg cos ^2 theta in the y

how does that make fr = mgsinθcosθ x+ mgsin^2y

 

[Tinhorn]

Can i say mgsinθ is the force moving the box down the slope

since the x and y components of the force A is
$A_{x}$= A*cosθ
$A_{y}$= A*sinθ

I can say mg sin θ = mg sinθcosθ + $mgsin^{2}$

and since weight or gravity is also working on the box
which is -mg on the y component.

dot product would be

$mgsinθcosθ _{x}$- $mgsin^{2} _{y}$
which is the resultant (i copied the sign wrong last time)

 

[Doc Al]

Can i say mgsinθ is the force moving the box down the slope

since the x and y components of the force A is
$A_{x}$= A*cosθ
$A_{y}$= A*sinθ

I can say mg sin θ = mg sinθcosθ + $mgsin^{2}$

and since weight or gravity is also working on the box
which is -mg on the y component.

dot product would be

$mgsinθcosθ _{x}$- $mgsin^{2} _{y}$
which is the resultant (i copied the sign wrong last time)

That's the correct resultant, but I don't follow your reasoning. Here's how you get it by adding the force components you found earlier:

Add the x components (there's only one): mgcosθsinθ
Add the y components: mgcos²θ - mg = -mgsin²θ

 

[Tinhorn]

I get it.
but i had always thought the x and y components of the force A is
Ax= A*cosθ
Ay= A*sinθ

why does normal force have sin in its component
and cos in its component