What is the Maximum Energy for a Child on a Pogo Stick?

[AznBoi]

Homework Statement

A child's pogo stick stores energy in a spring (k=2.5 x 10^4 N/m). At position A (x_1=-0.1m), the spring compression is a maximum and the child is momentarily at rest. At position B (x=0), the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. There is a displacement (x_2) above the x=0 mark. Assuming that the combined mass of the child and pogo stick is 25kg, a) Calculate the total energy of the system if both potential energies are zero at x=0, b) determine x_2, c) calculate the speed of the child at x=0, d) determine the value of x for which the kinetic energy of the system is a maximum, and e) obtain the child's maximum upward speed.

Homework Equations

E(before)=E(after)--- Both Potential and Kinetic

The Attempt at a Solution

I wasn't able to find a) right off the bat. The answer I got (125J) does not match the answer in the back of the book (101J). What I did to find a) was put: U(spring)= KE(after) , and to find U(spring) I used:
1/2kx^2 --> 1/2(2.5 x 10^4 N/m)(-0.1m)^2

That should give you 125J but again, it doesn't match the answer given. Am I doing something wrong? Something about the displacement of the spring?

I got b) right I think. To find x_2 I used KE(before)=U_g(after), so:
1/2mv^2=mgh ---> 101J(Used book's answer)= (25)(9.8)h

I solved for h and i got .41m, which is supposedly correct. =]

I got c) by using U(spring)=KE(after) so: 101J=1/2(25)v^2; v=2.8425

So overall, I don't know how to find a), d) or e).

For both d) and e), how do you know where the KE and speed is maximum?

 

[AlephZero]

Think about the potential energy of the child and stick (relative to the datum position x = 0) as well as the potential energy in the spring.

For (d) and (e), the KE+PE is constant, so the KE is a maximum when the PE is a minumum. You can write down the PE as a function of x.

 

[AznBoi]

Okay, I get how KE+PE is always constant. But how do you mean either one a maximum/minimum?? make it equal to .00000000000001 or something? Can you show me how to make it "a function of x"?

 

[radou]

Okay, I get how KE+PE is always constant. But how do you mean either one a maximum/minimum??

The answer is in your question, since, if the sum is constant, minimizing one summand maximized the other, and vice versa.

 

[AznBoi]

The answer is in your question, since, if the sum is constant, minimizing one summand maximized the other, and vice versa.

d) determine the value of x for which the kinetic energy of the system is a maximum, and e) obtain the child's maximum upward speed.

You need to find numbers. For d) its -9.8mm and e) 2.85 m/s?

How do you find those exactly? I'm going to try something.

 

[Saketh]

Using conservation of energy, you can find the answer.

We can help you more easily if you show us your conservation of energy expression. Once you figure out how to write that, the rest is just arithmetic.

 

[AlephZero]

When the pogo stick is below x = 0, the PE is (1/2)Kx^2 - mgx (you left out the mass term in your attempt at the solution). At the minimum (or maximum), d(PE)/dx = 0.

 

[AznBoi]

When the pogo stick is below x = 0, the PE is (1/2)Kx^2 - mgx (you left out the mass term in your attempt at the solution). At the minimum (or maximum), d(PE)/dx = 0.

Ohh.. alright I got the first part. So any time x is not zero you have to use the gravitational potential energy also? I thought the only potential energy it had was from the spring. Anyways.. Isn't the total energy in the system always the same? Unless it is nonconservative... Because if you make the potential energy 0 the Kinetic energy would have 101J, the same as potential energy before. What am I doing wrong? Thanks. Also, I don't know how to use calculus or any of those derivations.

 

[Saketh]

Ohh.. alright I got the first part. So any time x is not zero you have to use the gravitational potential energy also? I thought the only potential energy it had was from the spring. Anyways.. Isn't the total energy in the system always the same? Unless it is nonconservative... Because if you make the potential energy 0 the Kinetic energy would have 101J, the same as potential energy before. What am I doing wrong? Thanks. Also, I don't know how to use calculus or any of those derivations.

Potential energy is completely relative. Gravitational potential energy (our mgh approximation) has no meaning except in terms of a certain reference point with zero gravitational potential.

Let's set the point where the spring is fully compressed as zero gravitational potential energy. You can also set it to any other point - but you have to be able to write the gravitational potential energy expression. They'll all come out to be the same thing in the end, regardless of where you have your reference point.

Anyways.. Isn't the total energy in the system always the same?

In this particular system, yes.

But we have to take into account gravitational potential energy, like this:

$$\left (PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right)_{initial} = \left ( PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right )_{final}$$

Why can we say this? Let's think about it. Energy is conserved - as you said - so all the initial forms of energy must equal the final forms of energy. It remains to be seen which ones are zero.

When the spring is fully compressed, there is just spring potential energy. So we write that down. Let's take the fully compressed point to be our zero gravitational potential energy point. Also, when the spring is fully compressed, the stick and the boy aren't moving - no initial kinetic energy. Let's strike those terms, then.

$$\left (PE_\rmmath{spring})_{initial} = \left ( PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right )_{final}$$

Now it's just a matter of deciding what happens next. Is your final state the one where the spring is fully decompressed? If so, there's no spring potential energy in the end. Or is it the one where the stick and boy have reached the maximum height? Then there's no kinetic energy or spring energy on the right side.

In summary, here's how you want to approach conservation of energy problems:

1. Determine if there is any nonconservative work being done. If so, take that into account (that's not the case here).
2. Write out all of the possible forms of energy as "initial" and "final".
3. Determine what forms of energy exist at the beginning. Strike out those that are zero.
4. Determine what forms of energy are there in the final state. Strike out the ones that are zero.
5. Solve for whatever you are solving for.

Be careful! It's tricky to get the gravitational potential energy expressions correct.

 

[AznBoi]

Thanks for the awsome reply, but I'm still confused about a) calculate the total energy of the system if both potential energies are zero at x=0.. The equation that I used is 1/2kx^2+mgh=K, I just solved for the left side and I got 101J, which is correct as state in the back of the book. Why do you need to consider the mgh??

or mabye I wrote it wrong. it should be 1/2kx^2=mgh right? because at first the spring is fully compressed, then when the child reaches his max height, it is gravitational potential energy?

I still don't get how you would solve for the maximum of something, do you just make something 0?

 

[Saketh]

Thanks for the awsome reply, but I'm still confused about a) calculate the total energy of the system if both potential energies are zero at x=0.. The equation that I used is 1/2kx^2+mgh=K, I just solved for the left side and I got 101J, which is correct as state in the back of the book. Why do you need to consider the mgh??

or mabye I wrote it wrong. it should be 1/2kx^2=mgh right? because at first the spring is fully compressed, then when the child reaches his max height, it is gravitational potential energy?

I still don't get how you would solve for the maximum of something, do you just make something 0?

They're telling you that you should use x_0 as the zero gravitational potential energy point.

So the total energy would be all potential when the spring is fully compressed, which is equal to:

$$PE_{grav} + PE_{spring}$$

Right?

So now we have
$$PE_{grav} + 125J$$

Gravitational potential energy is negative, because we are below the reference point, so that would be -mgh, which is -24.5J.

I still don't get how you would solve for the maximum of something, do you just make something 0?

No - you have to think about what becomes zero (from the complete initial-final energy expression) when something is at a maximum. There's a lot of thinking in these energy problems, and once you're done thinking the rest is just arithmetic.

(EDIT: I sent you a PM about the other thing.)

 

[AznBoi]

Hmm, well if the energy is conserved and all the energy before is the same as after. U(before)=K(after) if U was 101J before, then wouldn't K(after) be 101J if there was no potential energy left and all of it was K?

Thanks a lot for all your help btw. You pm too =D

 

[Saketh]

Hmm, well if the energy is conserved and all the energy before is the same as after. U(before)=K(after) if U was 101J before, then wouldn't K(after) be 101J if there was no potential energy left and all of it was K?

Thanks a lot for all your help btw. You pm too =D

No, because all of it will never be KE.

Think about what I just said, and how it can be true. (Hint: Gravitational potential energy.)

 

[AznBoi]

hmm I think I have the equation:

1/2kx^2+mgh=101J

x and h are supposedly the same. To get the x of max KE, I just solve for x right?

 

[Saketh]

No, you would want

1/2 kx^2 + mgx_0 = KE_max + mgx = 101J

where x is the distance above the x=0 line. Do what AlephZero said and differentiate.

To find maximum velocity, use the above equation.

 

[AlephZero]

Also, I don't know how to use calculus or any of those derivations.

Hmm... I can't think of a way to do part (d) without using calculus - or at least, without using the same general ideas you would learn in a first calculus course. I can't imagine you are supposed to work out those ideas for yourself, without taking the course - and this isn't the right place to write a textbook about calculus...

 

[AznBoi]

wow really?? yeah I got only the first few a) b) c), but I can't get d) or c) I tried using the equation but I end up with .55 m which is 550mm, the book says -9.8mm =P I don't know what I'm doing wrong... Seriously, you need calculus to solve this??

 

[Doc Al]

Seriously, you need calculus to solve this??

You don't need calculus to do part d (or e). Hint: Consider the net force on the system as the spring decompresses.

 

[AznBoi]

So for d) determine the balue of x for which the kinetic energy of the system is maximum..

Do I use PE=KE and just substitute all of the correct numbers in?

 

[Doc Al]

d)

Do I use PE=KE and just substitute all of the correct numbers in?

No. You need to determine the position at which the KE is maximum. Do that by considering the net force on the system as it moves from its lowest point. Hint: What will the net force be when KE is maximum?

 

[AznBoi]

No. You need to determine the position at which the KE is maximum. Do that by considering the net force on the system as it moves from its lowest point. Hint: What will the net force be when KE is maximum?

The net force would be pointed upwards right? Wait, how does the net force relate to the position of the pogo stick? Do you have to use Energy(before)=Energy(after) to solve it?

 

[Doc Al]

d) cont.

The net force would be pointed upwards right? Wait, how does the net force relate to the position of the pogo stick?

Only two forces act on this system. What are they? Consider the net force on the system, starting from the lowest point. Clearly, that net force must point upwards, otherwise the system would not move upward. And, as long as the net force is upward, the system will accelerate upward, going faster and faster. At what point will it stop accelerating upward? (What will the net force be at that point? Solve for that point!)

Do you have to use Energy(before)=Energy(after) to solve it?

No!

 

[AznBoi]

Only two forces act on this system. What are they? Consider the net force on the system, starting from the lowest point. Clearly, that net force must point upwards, otherwise the system would not move upward. And, as long as the net force is upward, the system will accelerate upward, going faster and faster. At what point will it stop accelerating upward? (What will the net force be at that point? Solve for that point!)


No!

Well, at first I thought the pogo stick accelerated upwards the fastest at the middle point (between take off and the highest point where v=0)

I think it is the middle because PE goes to KE then back to PE and won't the pogo stick start losing speed after halfway?

I don't get what equation you would use to solve for the distance/velocity of the max KE.

 

[Doc Al]

Well, at first I thought the pogo stick accelerated upwards the fastest at the middle point (between take off and the highest point where v=0)

I think it is the middle because PE goes to KE then back to PE and won't the pogo stick start losing speed after halfway?

Why the middle? What physical fact describes the point at which the speed is greatest. Big Hint: As long as the pogo stick is accelerating upwards it can't be moving the fastest, right? (Since an instant later it will be going even faster.) So what you want is the moment when the acceleration stops--what will the net force be then?

I don't get what equation you would use to solve for the distance/velocity of the max KE.

First find the position by considering the net force. Once you have the position, you can use energy conservation to find the KE.

 

[AznBoi]

Why the middle? What physical fact describes the point at which the speed is greatest. Big Hint: As long as the pogo stick is accelerating upwards it can't be moving the fastest, right? (Since an instant later it will be going even faster.) So what you want is the moment when the acceleration stops--what will the net force be then?


First find the position by considering the net force. Once you have the position, you can use energy conservation to find the KE.

Oh, so the pogo stick will be moving the fastest right after take off. Then it was start decelerating?

When the upward acceleration stops, the net force would be pointed down?
Wait, is when the acceleration stops at the top? The net force would be down also.

If we are trying to find the max KE distance, surely it wouldn't be the top point right?

 

[Doc Al]

Another hint: What does Newton's 2nd law say about the relationship between acceleration and net force?

 

[AznBoi]

Another hint: What does Newton's 2nd law say about the relationship between acceleration and net force?

That acceleration is directly proportional to the net force. So that means for the acceleration to equal zero, either the mass or the net force has to equal zero. However, the mass is not equal to zero in this case so that means the net force has to be equal to zero in order to have 0 acceleration?

So the object cannot be at the top because it would have a downward net force from gravity.

Is there still a force going up after the pogo stick has left the bottom? I think I'm getting confused about velocity/accel vectors and forces(Newtons). Weight is always a force, correct? Is there a force going upwards? What force is that? I mean it is caused by the potential energy in the spring but after that what do you call the velocity/acclerationg vecotr going upwards?

Does that mean the max KE is right before a=0, net force=0? How could you calculate this and find the distance?

 

[Doc Al]

That acceleration is directly proportional to the net force. So that means for the acceleration to equal zero, either the mass or the net force has to equal zero. However, the mass is not equal to zero in this case so that means the net force has to be equal to zero in order to have 0 acceleration?

That is correct.

So the object cannot be at the top because it would have a downward net force from gravity.

Not sure what you mean here, but it's certainly true that at the top of the motion the only force on the system is gravity, which acts downward.

Is there still a force going up after the pogo stick has left the bottom? I think I'm getting confused about velocity/accel vectors and forces(Newtons). Weight is always a force, correct? Is there a force going upwards? What force is that? I mean it is caused by the potential energy in the spring but after that what do you call the velocity/acclerationg vecotr going upwards?

In this problem there are only two forces acting:
(1) The spring force, which is proportional to the amount of stretch in the spring and which acts upward;
(2) Gravity, which is constant and acts downward.

The net force is the combination of those two forces.

Does that mean the max KE is right before a=0, net force=0? How could you calculate this and find the distance?

When the upward acceleration is zero (and thus the net force = 0), the pogo stick is going as fast as it's going to go, thus it has maximum KE. (As it continues upward, it will only slow down since the net force will be downward.)

You need to solve for the point where the net force equals zero.

 

[AznBoi]

I'm truly lost here. I don't know how you could use the net force information to calculate the max KE. Can you tell me an equation I can use to calculate it? I tried making the Potential energy of the sprign and gravitational potential energy equal 0 but I came up with a number that was off by not much, but I still know it is wrong.

I did: mgh+1/2kx^2=0 because I figured if it was all KE then there would be no potential energy. I don't get it, please help.

 

[Doc Al]

Not sure what more I can say. Every time I tell you to find the net force, you start pulling out energy equations again.

You know the two forces, so you should be able to find the net force as a function of the position below the unstretched position of the pogo spring. Do that!

You know (or at least I've stated) that the position of max KE will occur when the acceleration (and thus the net force) equals zero*. So you should be able to write an equation and solve for the position at which net force = zero.

(*If that statement still puzzles you, reread my earlier posts until it clicks.)

 

[AznBoi]

Yay, I found the position by doing what you said lol. I realized that the spring force has distance in it. F(spring)+W=0 --> -kx=mg and x=-9.8x10^-3

However, I was unable to find the velocity of the max KE. Do you just plug x into 1/2kx^2+mgx=K and solve for v? I did that but it doesn't work.. Something is weird here though, the velocity is only .01 m/s greater than x=0.1m

 

[AznBoi]

What I did to find the velocity of the max KE was:

101J=1/2mv^2+mgx+1/2kx^2

I plugged -.0098m for x and I solved for v I got 2.8591m/s but the answer says 2.85 m/s. I don't know if I did this correctly because I don't think 5 can be rounded down? I never learned about sig figures and I think I've heard something about 5 being rounded down.

Did I solve the velocity of max KE correctly? I considered all of the potential energies because x is not 0...

 

[Doc Al]

Sorry for the delayed response... I forgot to follow up on this thread.

Yay, I found the position by doing what you said lol. I realized that the spring force has distance in it. F(spring)+W=0 --> -kx=mg and x=-9.8x10^-3

Good.

What I did to find the velocity of the max KE was:

101J=1/2mv^2+mgx+1/2kx^2

I plugged -.0098m for x and I solved for v I got 2.8591m/s but the answer says 2.85 m/s. I don't know if I did this correctly because I don't think 5 can be rounded down? I never learned about sig figures and I think I've heard something about 5 being rounded down.

Did I solve the velocity of max KE correctly? I considered all of the potential energies because x is not 0...

You solved it correctly. I wouldn't worry about the 3rd sig figure, since the initial data was only given to 2 sig figures anyway.