Spring Launching a Block

  • #1

Homework Statement


A spring that has a spring constant of 2000 N/m is compressed 5 centimeters from its rest position. If a 2.5kg block is placed on the edge of the spring and then released, what will the block's kinetic energy be once it has completely left the spring? How fast will the block be moving?

Spring Constant: 2000 N/m
Spring Compression: 5cm
Block Mass: 2.5kg
Assumed Gravity Force: 10m/s^2

Homework Equations


KE = 1/2mv^2
SpringForce = (SpringConstant)(SpringCompression)

The Attempt at a Solution


I entered in the values of the spring constant to find the force of the spring, multiplying 2000 by 5, getting an answer of 10,000N for the force of the spring. Then I moved to put this information into the second equation, for kinetic energy, and realized that I'm missing something in between. The two questions that are included in this problem actually go together. Once I find one, I should be able to find the other (I think). To find the kinetic energy of the block however, don't I need the velocity? And vice versa? Thanks in advance!
 

Answers and Replies

  • #2
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The usual approach to this problem is to first equate the stored energy of the cocked spring to the launch energy of the block, assuming that the spring has come to rest as the block separates. (This is somewhat questionable. How can the spring be coming to rest when the block has received all the stored energy and has max velocity?) Once the launch energy of the block is known, then the velocity is easily obtained.

A more correct approach might be to track the contact force between the spring and the block as the block is moved by the spring. When the contact force goes to zero, the block is launched and the energy at that point is the launch energy.
 
  • #3
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172
Are there any other equations in your text book or that your teacher covered in class that may be helpful for this problem? Since the force varies with the compression of the spring, you cannot simply use force x distance to calculate how much work the spring does on the block. But you could integrate it over the 5 centimeter distance. (Note that it is 5 cm, not 5 m that you used in your calculation of 10,000 N.)
 
  • #4
The usual approach to this problem is to first equate the stored energy of the cocked spring to the launch energy of the block, assuming that the spring has come to rest as the block separates. (This is somewhat questionable. How can the spring be coming to rest when the block has received all the stored energy and has max velocity?) Once the launch energy of the block is known, then the velocity is easily obtained.

A more correct approach might be to track the contact force between the spring and the block as the block is moved by the spring. When the contact force goes to zero, the block is launched and the energy at that point is the launch energy.
Both of those statements make some sense to me...my apologies, I'm not very good at physics...is there a way to determine when the contact force will be zero? I mean, it's going to be when the spring is fully extended, theoretically.
 
  • #5
Are there any other equations in your text book or that your teacher covered in class that may be helpful for this problem? Since the force varies with the compression of the spring, you cannot simply use force x distance to calculate how much work the spring does on the block. But you could integrate it over the 5 centimeter distance. (Note that it is 5 cm, not 5 m that you used in your calculation of 10,000 N.)
My professor didn't exactly cover springs in class...he's using the homework from past professors but doing his own lesson plan, so the homework doesn't always line up with what he taught in class. What I know about springs is solely from Google, sadly. :/
 
  • #6
686
172
What I know about springs is solely from Google, sadly. :/
If you were to do a Google search of "what is the stored energy of a compressed spring", you would find that it is equal to (1/2)kx2.
 
  • #7
If you were to do a Google search of "what is the stored energy of a compressed spring", you would find that it is equal to (1/2)kx2.
Ah, okay. And since there is no friction, I can assume that the potential energy of the spring will equal the kinetic energy of the block, correct? So the PE in the spring is 25,000J then, based on (1/2)(2,000)(52). Which would make the Kinetic Energy as the PE left the spring equal to 25,000J as well, meaning that the KE of the block is 25,000J and the velocity is the square root of 20,000, so roughly 141cm/s, which I could then convert into 1.41m/s if I wanted a more correct answer. Is this the right way to do it?
 
  • #8
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172
Is this the right way to do it?
You are mixing units in your calculation of energy. The spring constant is N/m and you are plugging 5 cm into the equation. The incorrect result is 25,000 J. You should convert 5 cm to 0.05 meters before you plug it in, which will result in 2.5 J. Except for that, it looks right.
 
  • #9
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To build off of @TomHart's response, make sure you always have your numbers in the correct SI units before solving (unless you are asked to solve for different units).
 
  • #10
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It turns out that the point where contact is lost is when the spring is again at its free length. At any previous point, there would still be compression in the spring and therefore some amount of contact force.

With the numbers given by the OP, I get a launch velocity of 1.4142 m/s, assuming that it all takes place on a frictionless horizontal plane.

What is this stuff about "Assumed Gravity Force: 10m/s^2"? If this is intended to represent g, the acceleration of gravity, this is an exceptionally poor approximation to 9.807 m/s^2. Fortunately, it is irrelevant, at least as I see it.
 
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  • #11
haruspex
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With the numbers given by the OP, I get a launch velocity of 10.6066 m/s
I confirm √2 m/s.
This is somewhat questionable. How can the spring be coming to rest when the block has received all the stored energy and has max velocity?)
I do not see the difficulty. The spring is presumed massless, so has no inertia.
 
  • #12
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I have corrected my previously erroneous velocity value above and agree with haruspex on this.

My speculation about the contact loss occuring at the free length was corrected in #10, so I think haruspex may have overlooked this.

As it turns out, even if an equivalent mass for the spring is included (spring mass/3) in the formulation, the loss of contact still occurs when the spring reaches its free length.
 
  • #13
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As it turns out, even if an equivalent mass for the spring is included (spring mass/3) in the formulation, the loss of contact still occurs when the spring reaches its free length.
That's what I concluded also because that is the point where the spring force (and therefore, acceleration of the spring) reverses direction - that is, for a spring that has mass. When I say "acceleration of the spring", I'm talking about the acceleration of a point on the spring relative to the fixed end of the spring. Obviously, the magnitude of acceleration of a particular point on the spring depends on its distance from the fixed end of the spring.
 
  • #14
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To build off of @TomHart's response, make sure you always have your numbers in the correct SI units before solving (unless you are asked to solve for different units).

I'd like to observe that all that is really required is to use a CONSISTENT set of units. This thread has shown how easily it is to get units incorrect, even when using nominally SI units (cm is not an SI unit, strictly speaking). On another thread, we have recently had quite a go around over mixing SI and US Customary units.

It is well established; incorrect units can send you off into never-never land without you realizing what has happened. CONSISTENCY is the key.
 
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  • #15
haruspex
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My speculation about the contact loss occuring at the free length was corrected in #10,
I was aware that posts 2 and 10 appeared in conflict, but since you did not explicitly state that you were withdrawing the earlier remark I thought there was room for confusion.
In particular, the wording in post 2 suggested to me you were concerned about inertia. I agree that the inertia doesn't matter anyway since the spring will expand uniformly. In general, though, massive springs can behave in a more complex manner, with internal oscillations.
 

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