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

In summary: Therefore, it should not include the x distance. In summary, the problem involves finding the distance the spring is compressed when the block comes to rest. The equation for potential energy is (1/2)kx^2 and the equation for kinetic energy is (1/2)mv^2. Using conservation of energy, the total energy in the initial position must equal the total energy in the final position, including potential gravitational and elastic energy. The solution involves solving for x, which represents the vertical distance the block moves.
  • #1
NewSoul
14
0

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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  • #2
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.
 
  • #3
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:
  • #4
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.
 
  • #5
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.
 

1. What is potential energy of a spring and mass on an inclined plane?

Potential energy of a spring and mass on an inclined plane is the energy stored within the spring-mass system when the spring is stretched or compressed and the mass is at a certain height on an inclined plane. This potential energy is a result of the elastic potential energy of the spring and the gravitational potential energy of the mass.

2. How is the potential energy of a spring and mass on an inclined plane calculated?

The potential energy of a spring and mass on an inclined plane can be calculated using the formula: PE = mgh + 0.5kx², where m is the mass, g is the acceleration due to gravity, h is the height of the mass on the inclined plane, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

3. What is the relationship between the potential energy of a spring and mass on an inclined plane and the angle of inclination?

The potential energy of a spring and mass on an inclined plane is directly proportional to the angle of inclination. This means that as the angle of inclination increases, the potential energy also increases.

4. How does the mass of the object affect the potential energy of a spring and mass on an inclined plane?

The potential energy of a spring and mass on an inclined plane is directly proportional to the mass of the object. This means that as the mass increases, the potential energy also increases.

5. Does the potential energy of a spring and mass on an inclined plane change if the object is moving?

Yes, the potential energy of a spring and mass on an inclined plane changes if the object is moving. This is because the potential energy is also dependent on the position of the object, and if the object is moving, its position and height on the inclined plane will change, thereby changing its potential energy.

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