Conservation of energy (Ramp question) - Thanks

In summary: But instead of assuming that the mass had reached its maximum height, you were assuming that it had reached the bottom of the ramp, this is why you were getting a meaningless answer. Your answer was the maximum height minus the initial height, which was (1/2)kx^2/mg. This is nonsense, because it doesn't even tell you what h is! It's just some number. So this is the only mistake you made. But you had the right idea. In summary, the problem involves using conservation of energy to find the height that a mass will reach when released from a compressed spring at the top of a ramp with a frictionless flat portion. The initial energy is all in the form of elastic potential energy,
  • #1
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Homework Statement



Here is the question I am having issues with... Below are my questions regarding it.

9ap7kh.jpg



Homework Equations





The Attempt at a Solution



Now when looking at this, first I would assume I use conservation of momentum.

(K + Ug + Us)i = (K + Ug + Us)f + Ff d

Now for the first part (left side of the equation) my text says the following.

0 + 0 + 1/2Kx^2

I don't get that. So when its at the start, There is no Kinetic energy and no potential evergy, and for the spring energy its just 1/2Kx^2 ?

I don't understand how to figure out the right side of that equation, and mainly finding the height, as outlined in the question.

THanks!
 
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  • #2
nukeman said:

Homework Statement



Here is the question I am having issues with... Below are my questions regarding it.

9ap7kh.jpg

Homework Equations


The Attempt at a Solution



Now when looking at this, first I would assume I use conservation of [STRIKE]momentum[/STRIKE] energy.

(K + Ug + Us)i = (K + Ug + Us)f + Ff d

This is correct, once you include my correction above (which I'm sure was just a typo anyway).

nukeman said:
Now for the first part (left side of the equation) my text says the following.

0 + 0 + 1/2Kx^2

I don't get that. So when its at the start, There is no Kinetic energy and no potential evergy, and for the spring energy its just 1/2Kx^2 ?

Yes. There is no kinetic energy because the mass is being held *at rest* against the spring. There is no gravitational potential energy, because the mass is at h = 0. The only energy in the system initially is the elastic potential energy stored in the spring, which is indeed given by (1/2)kx2, where x is the displacement of the spring from its equilibrium (unstretched/uncompressed) position. That's really all there is to it. So, if you're still confused, you'll have to elaborate about what specifically is confusing you.

nukeman said:
I don't understand how to figure out the right side of that equation, and mainly finding the height, as outlined in the question.

THanks!

The mass will be released from the spring at the moment that the spring is entirely uncompressed (i.e. x = 0), which means that all of the elastic potential energy is gone. But energy is conserved, which means that all of the elastic potential energy stored must have been converted into kinetic energy of the mass during the release. So, due to conservation of energy, you know how much KE the mass has upon release. Furthermore, the flat portion has no friction, which means that the mass still has this amount of kinetic energy when it reaches the base of the ramp. Therefore, you can use conservation of energy a second time (taking into account work done by friction as well) to determine what height the mass will reach.
 
  • #3
I just realized it's not necessary to do it as a two step process though. You could apply conservation of energy only once, just as you were planning to, and you could use the fact that the kinetic energy is also zero at the end (i.e. on the right hand side of the equation). So you have (1/2)kx^2 = mgh +W_friction

Why is the kinetic energy 0 on the right-hand side as well? It's because the mass has reached its *maximum* height up the ramp, therefore it must be stopped. If it were still moving, then it would go up farther.

So you had the basic idea all along.
 

1. What is the principle of conservation of energy?

The principle of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another.

2. How does the ramp experiment demonstrate the conservation of energy?

In a ramp experiment, a ball is rolled down a ramp and then up another ramp of the same height. The ball reaches the same height on the second ramp, demonstrating that the potential energy at the top of the first ramp is equal to the kinetic energy at the bottom of the ramp and the potential energy at the top of the second ramp.

3. What is the equation for calculating potential energy?

The equation for potential energy is PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.

4. How does friction affect the conservation of energy in a ramp experiment?

Friction can slightly decrease the kinetic energy of the ball as it rolls down the ramp, but it does not change the overall conservation of energy principle. Some of the ball's potential energy is converted to thermal energy due to friction, but the total amount of energy remains the same.

5. Can the conservation of energy be applied to other situations besides the ramp experiment?

Yes, the principle of conservation of energy can be applied to many different situations, such as a pendulum swinging, a roller coaster ride, or a battery powering a lightbulb. In all of these cases, the total amount of energy remains constant, but it may be transformed into different forms.

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