Maximum speed of object between 2 springs

In summary, the maximum speed of the 1.5 kg box moving between two springs on a frictionless surface is 1.85 m/s. This was calculated using the conservation of energy rule and setting the initial and final kinetic energy and elastic potential energy equal to each other. The final kinetic energy was found by subtracting the final elastic potential energy and dividing by the mass, and then taking the square root. It was important to only include one of the potential energy terms in the equation, as the other spring is not compressed when the block has potential energy in one spring.
  • #1
rolodexx
14
0
[SOLVED] Maximum speed of object between 2 springs

Homework Statement



A 1.50 kg box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 cm, and then is released from rest.
(I solved part A, B is: What is the maximum speed the box will reach?)

Homework Equations


K[tex]_{i}[/tex] + EPE[tex]_{i}[/tex] = K[tex]_{f}[/tex] + EPE[tex]_{f}[/tex]

.5mv[tex]^{2}[/tex] [tex]_{i}[/tex] + .5k[tex]_{i}[/tex]x[tex]_{i}[/tex] [tex]^{2}[/tex] = .5mv[tex]^{2}[/tex][tex]_{f}[/tex] + .5k[tex]_{f}[/tex]x[tex]^{2}[/tex][tex]_{f}[/tex]

(sorry about the messed-up subscripts; I can't make it work right)

The Attempt at a Solution


I solved for x[tex]_{}f[/tex] first, obtaining the correct answer of 5.66 cm.

I then plugged in values for the other variables: m of 1.5 kg, initial velocity of 0 (because it started from rest?), initial k of 32 N/m, initial x of 4 cm, final k of 16 N/m, final x of 5.66 cm. I converted cm to m and tried to solve for final velocity, but I got .062 m/s and it's supposed to be 1.85 (I got it wrong enough times that it gave me the answer, I just don't know how to solve for it).
 
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  • #2
I am not sure exactly how you are trying to solve for the maximum velocity. Can you show your calculations and reasoning? I can't help find your mistake if I can't see it.
 
  • #3
v[tex]_{f2}[/tex] = mv[tex]_{i}[/tex][tex]^{2}[/tex] + k[tex]_{i}[/tex]x[tex]_{i}[/tex][tex]^{2}[/tex] - k[tex]_{f}[/tex]x[tex]_{f}[/tex][tex]^{2}[/tex]

I used the conservation of energy rule to set the initial and final kinetic energy and elastic potential energy equal to each other, then tried to solve for final kinetic energy by subtracting both sides by final elastic potential energy. Then I divided the final kinetic energy by the mass to get final velocity squared. On the other side, that left initial velocity squared (its mass canceled out when I divided by the mass from the other side) plus initial spring constant * inital compression, minus final spring constant * final compression
After I plugged in the numerical values stated above, I took the square root to change final velocity squared to just final velocity.
 
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  • #4
Your problem is that you have both the potential energy terms involved.

When the block has potential energy in one spring, the other spring is not compressed at all. So, the total energy is either:

[tex]TE=1/2k_ix_i^2[/tex] or [tex]TE=1/2k_fx_f^2[/tex]

Thus, only one of the above should be involved in the equation you posted above.
 
  • #5
OH! I feel stupid now... lol. I'll try that and see if I can make my problem work! Thanks.
 
  • #6
The assumption I was relying on was initial velocity is 0... since there were no values aside from initial compression that could be used to get velocity. Is this a correct assumption? Because with a zero velocity, initial kinetic energy is also 0, so that would leave me with final kinetic energy equal to initial elastic potential energy. So initial EPE divided by mass gives final v^2?

edit: oh! I got it right... my last fatal mistake was a unit conversion of cm to m. lol. But I made it! Thank you again for your help.
 
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  • #7
Anytime.:smile:
 

1. What is the maximum speed of an object between 2 springs?

The maximum speed of an object between 2 springs is determined by the properties of the springs, such as their stiffness and length, as well as the mass of the object. This speed can be calculated using the equation v = √(k/m), where k is the spring constant and m is the mass of the object.

2. How does the stiffness of the springs affect the maximum speed of the object?

The stiffness of the springs has a direct impact on the maximum speed of the object. The higher the stiffness, the greater the force exerted by the springs and the faster the object will move between them.

3. Can the maximum speed of the object be greater than the speed of sound?

Yes, the maximum speed of the object can be greater than the speed of sound. This is because the speed of sound is determined by the properties of the medium, while the maximum speed of an object between 2 springs is determined by the properties of the springs and the object, which can vary greatly.

4. Is the maximum speed of the object affected by the distance between the two springs?

Yes, the maximum speed of the object is affected by the distance between the two springs. As the distance increases, the force exerted by the springs decreases, resulting in a lower maximum speed.

5. How can the maximum speed of the object be increased?

The maximum speed of the object can be increased by increasing the stiffness of the springs, decreasing the mass of the object, or decreasing the distance between the two springs. Additionally, adding more springs in series or in parallel can also increase the maximum speed of the object.

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