Maximum compression of a spring

In summary, the problem asks for the maximum compression of a spring when a 50-kg mass is dropped from a 20-m height onto a spring with a force constant of 100 N/m. Using the conservation of energy and assuming the equilibrium position does not affect the compression of the spring, the final equation is Ugo=Uef+Ugf, where Ugo is the initial gravitational potential energy, Uef is the final elastic potential energy, and Ugf is the final gravitational potential energy. By measuring gravitational potential energy with the lowest position of the mass as the zero point, the final position will only have elastic potential energy. The amount of compression can be represented as X. The final equation can then be
  • #1
supersingh
9
0

Homework Statement




A 50-kg mass is dropped from a 20-m height onto a very long, initially uncompressed spring of force constant k = 100 N/m

Homework Equations


Ueo=Ugf+Uef
F=k(Lenght Final-length initial)


The Attempt at a Solution


I tried using Ueo=Ugf+Uef, and i couldn't figure out what to use for some of the values.
so far i got 1/2*4.9^2=Ugf+Uef, i don't know what to put for the final heights.
 
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  • #2
supersingh said:

Homework Statement




A 50-kg mass is dropped from a 20-m height onto a very long, initially uncompressed spring of force constant k = 100 N/m

Em but what exactly is the question asking to solve here?
 
  • #3
oh my bad, its asking what is the maximum compression of the spring
 
  • #4
supersingh said:
oh my bad, its asking what is the maximum compression of the spring

Ahh my bad, should have heeded the title a little better.:blushing:

Well, what will determine how much the spring decreases in length by? Using this can you work out the length the spring decreases by?
 
  • #5
i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50
 
  • #6
supersingh said:
i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50
The equilibrium point is not needed for this problem.

Hint: In calculating the change in gravitational PE, be sure to use the total change in height.
 
  • #7
supersingh said:
i found out that when the object is in equllibrium, the spring goes displaces 4.9 m using
mg=k(l-L) where k is 100 and the mass is 50

Yes but are you sure it's [mg=k(l-L)]

Edit: Ahh I've been beaten me here.
 
  • #8
im pretty sure it is, becuase the problem before it needed that height to calculate the velocity at equillibrium, and i got that
 
  • #9
supersingh said:
im pretty sure it is, becuase the problem before it needed that height to calculate the velocity at equillibrium, and i got that

Maybe, but in this case the equilibrium position doesn't matter, because the most the spring can be compressed is the force of mass that fell onto it.
 
  • #10
Another hint: This is a conservation of energy problem.
 
  • #11
in the intial situations, there is only intial elastic potential
then for the final, there is final elastic and final gravitational. what value should i use for the height for the gravitational value?
 
  • #12
Initially, there's only gravitational PE. Hint: If you measure gravitational PE using the lowest position of the mass as the zero point, then in the final position (of maximum compression) there will only be elastic PE.

Another hint: Call the amount by which the spring compresses X.
 
  • #13
wait, i got it, thanks everyone, i now used Ugo=Uef+Ugf
 

FAQ: Maximum compression of a spring

What is maximum compression of a spring?

Maximum compression of a spring refers to the point at which the spring has been compressed to its fullest extent and can no longer be compressed any further. This is also known as the spring's elastic limit.

How is maximum compression of a spring calculated?

The maximum compression of a spring can be calculated using Hooke's Law, which states that the force required to compress a spring is directly proportional to the distance it is compressed. The formula for this is F = -kx, where F is the force, k is the spring constant, and x is the distance the spring is compressed.

Why is maximum compression important?

Maximum compression is important because it allows us to determine the maximum amount of energy that can be stored in a spring. This is useful in various applications such as in shock absorbers, car suspension systems, and many other mechanical devices.

What factors can affect the maximum compression of a spring?

The maximum compression of a spring can be affected by several factors, including the material and size of the spring, the amount of force applied, and the temperature. Additionally, the spring's elastic limit can also be affected by repeated compression and relaxation cycles.

How can maximum compression be increased?

The maximum compression of a spring can be increased by increasing the force applied or by using a spring with a higher spring constant. Additionally, using multiple springs in series or in parallel can also increase the maximum compression. However, it is important to note that exceeding the spring's elastic limit can result in permanent damage or failure of the spring.

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