Solving Spring Constant: Find k with Hooke's Law

In summary, the problem involves a ball being dropped from a height of 50.0m above a spring and compressing it by 0.340m. Using conservation of energy and the equation for Hooke's law, the spring constant is calculated to be 38187.71 N/m.
  • #1
lking226
20
0

Homework Statement


A ball is dropped from rest at a height of 50.0m above a spring. After the ball hits, it compresses the spring .340m. Ignoring any non-conservative forces, find the spring constant if the mass of the ball is 4.50 kg.


Homework Equations


Hooke's law: k = mg/x


The Attempt at a Solution


I'm not sure what to do.
k = 4.50 kg x 9.8 m/s / 0.340 m = 129.71
That doesn't seem right.
Help?
 
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  • #2
Conservation of energy would work very well here.
 
  • #3
but how do you use conservation of energy with springs?
 
  • #4
lking226 said:
but how do you use conservation of energy with springs?

The potential energy stored in a spring is given by 1/2kx2. So what type of energy is being converted into the elastic potential energy in the spring?
 
  • #5
the kinetic energy of the ball?
 
  • #6
lking226 said:
the kinetic energy of the ball?

The energy is possesses at rest 50m above the spring.
 
  • #7
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?
 
  • #8
lking226 said:
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?

That would be correct. Can you now form an equation and solve for the spring constant 'k'?
 
  • #9
mghi + kxi^2 = mghf + kxf^2 ??
 
  • #10
lking226 said:
mghi + kxi^2 = mghf + kxf^2 ??

Initially the spring is not extended, so what does the left side reduce to? At the final stage, the final height is zero. So the entire equation simplifies to?
 
  • #11
so then just

mghi = kxf^2
 
  • #12
lking226 said:
so then just

mghi = kxf^2

Right, so what is k equal to now?
 
  • #13
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?
 
  • #14
lking226 said:
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?

That should be

[tex]4.5 kg \times 9.81 m/s^2 \times 50m = \frac{1}{2}k(0.34m)^2[/tex]


Solve again for k
 
  • #15
38187.71?
 
  • #16
lking226 said:
38187.71?

That looks correct to me
 
  • #17
Thank you so much!
 
  • #18
To nit-pick, the ball doesn't fall 50.0 meters, it falls 50.34 m after compressing the spring.
 

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force required to extend or compress a spring by some distance is directly proportional to that distance. This relationship can be represented by the equation F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.

2. How do you solve for the spring constant (k) in Hooke's Law?

To solve for the spring constant, you will need to know the force applied to the spring and the corresponding displacement from the equilibrium position. Once you have these values, you can use the formula k = -F/x to find the spring constant.

3. What units is the spring constant measured in?

The spring constant is typically measured in units of Newtons per meter (N/m). This unit represents the amount of force required to extend or compress the spring by one meter.

4. Can the spring constant change?

Yes, the spring constant can change depending on factors such as the material and dimensions of the spring. It can also change if the spring is stretched or compressed beyond its elastic limit.

5. How is Hooke's Law used in real-world applications?

Hooke's Law is used in various real-world applications, such as designing springs for different purposes (e.g. in mattresses, cars, or watches), studying the behavior of materials under stress, and even in medical devices like heart stents. It is also used in many scientific experiments to measure the properties of different materials.

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