Relating to potential energy of a spring and mass on inclined plane

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Homework Help Overview

The discussion centers around a physics problem involving the potential energy of a spring and a mass on an inclined plane. The scenario includes a block on an incline, a spring, and various parameters such as angle, mass, distance from the spring, and speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the conservation of energy principle, questioning the interpretation of energy terms and their relevance. There is discussion about the correct application of potential and kinetic energy equations, particularly in relation to the incline's angle and the spring's compression.

Discussion Status

Some participants have offered guidance on equating initial and final energy states, while others express confusion about the role of kinetic energy and the vertical height in the calculations. Multiple interpretations of the energy equations are being explored, and there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are grappling with the definitions and assumptions regarding potential energy, kinetic energy, and the effects of the incline on height calculations. There is uncertainty about how to incorporate the initial speed of the block into the energy equations.

NewSoul
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Homework Statement


Here is an illustration to ease understanding:
7-p-063.gif

Angle inclined plane: θ = 20.0°
spring of force constant k = 525 N/m
block: m = 2.71 kg
distance of block from spring: d = 0.330 m
speed of block: v = 0.750 m/s

By what distance is the spring compressed when the block momentarily comes to rest? (I'm going to call it 'x' in this problem.)

The answer should be x = 0.143 m

Homework Equations


Ok, I'm not quite sure how many of these are relevant or if I'm missing some, but I'll just give some that I know.

PEs = (1/2)kx2 (x = distance spring is compressed)
PEg = mgh
KE = (1/2)mv2
F = ma
U + KE = 0 (U = any potential energy? All potential energies? I'm not quite positive what is going on here.)

The Attempt at a Solution


Okay. I'm not quite sure how to go about this, so I'll just start doing what I seem to know.

Let's start with force. It doesn't seem inherently relevant to the problem, but why not?

Okay, so force in the x direction seems to be: mgsinθ = ma. So we get a = gsinθ = (9.8)(sin 20°) = 3.35198 m/s2.

Force in the y direction seems to be: N - mgsinθ = 0. So N = mgsinθ = (2.71)(9.8)(sin 20°) = 9.08337 N.

-------------

Ok. That was probably not relevant. Let's do work now.

Using the numbers above, I get KE = (1/2)(2.71)(.750)2 = 0.76219 J.

PEg = (2.71)(9.8)h. Okay, I'm not quite sure if h should just be d or if I have to factor in the slope. I'm thinking that slope probably does have to be included, so h is 0.330(sin 20°), so PEg is 2.99751 J.

So it seems like PEg + PEs + KE = 0. Then it must be that (1/2)kx2 + mgh + (1/2)mv2 = 0... However, solving for x gives me the square root of a negative number, I must be interpreting U + KE = 0 wrong. Where do I go from here? Thanks, I know this is long.
 
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Why do you think that the energy is zero?
You can use conservation of energy to solve the problem.
Equate the total energy in the initial position with the total energy in the final position.
Total energy here includes kinetic, potential gravitational and potential elastic.
Some terms may be zero, like elastic energy in the initial position or kinetic energy in the final position.
 
nasu said:
Why do you think that the energy is zero?
Well, I'm not entirely sure what is meant by the equation U + KE = 0 or if it's relevant...But I do know that a decrease in potential energy would mean an increase in kinetic energy and vice versa.

You can use conservation of energy to solve the problem.
Equate the total energy in the initial position with the total energy in the final position.
Total energy here includes kinetic, potential gravitational and potential elastic.
Some terms may be zero, like elastic energy in the initial position or kinetic energy in the final position.

So are you saying that PEg = PEs (KE is zero at the initial and final positions)? I'm afraid I don't understand.

*edit* Hold on a minute I did something weird...

Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?
 
Last edited:
NewSoul said:
Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?
h is the total vertical movement of the block. Don't forget that this will include a contribution from x.
 
NewSoul said:
So are you saying that PEg = PEs (KE is zero at the initial and final positions)? I'm afraid I don't understand.

*edit* Hold on a minute I did something weird...

Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?

No, the initial KE is not zero. It moves with some speed, isn't it?
And h is measured along the vertical direction.
 

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