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

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SUMMARY

The discussion focuses on calculating the distance a spring compresses when a block comes to rest on an inclined plane. Given parameters include an incline angle of θ = 20.0°, a spring constant k = 525 N/m, a block mass m = 2.71 kg, and an initial speed v = 0.750 m/s. The correct compression distance of the spring is determined to be x = 0.143 m, using conservation of energy principles. The participants clarify the relationship between potential energy, kinetic energy, and the spring's potential energy in this context.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with potential energy equations: PEs = (1/2)kx² and PEg = mgh
  • Knowledge of kinetic energy calculation: KE = (1/2)mv²
  • Basic trigonometry, particularly in relation to inclined planes
NEXT STEPS
  • Study the conservation of mechanical energy in systems involving springs and inclined planes
  • Learn how to derive and apply the equations for potential and kinetic energy
  • Explore the effects of angles on gravitational potential energy calculations
  • Investigate the role of initial velocity in energy conservation problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to energy conservation in spring-mass systems on inclined planes.

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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