Elastisity: Hooke's Law Finding Spring Compression

In summary, to find the amount of spring compression when a 28.86 kg child bounces on a pogo stick with a spring constant of 18016 N/m and experiences an upward acceleration of 4.802 m/s^2, you need to consider the total net force acting on the spring, which includes the force of gravity. Using the equation Fnet=([(mass given)*(acceleration given)] + [(mass given)*(gravity acceleration)])=kx, you can find that the spring is compressed by 0.007692369m.
  • #1
Prophet029
11
0
1) A 28.86 kg child bounces on a pogo stick. The pogo stick has a spring constant 18016 N/m. When the child makes a nice big bounce, she finds that at the bottom of the bounce she is accelerating upwards at 4.802 m/s2. How much is the spring compressed?

Given:
Mass of Child: 28.86kg
Spring constant 18016N/m
Acceleration after spring was compressed: 4.802 m/s^2

Find:
x=?

2) Relevant Equations

Hooke's Law: F=-kx
Newton's second Law: F=ma

3) Attempt at solution:

All I set up was that F=(Mass of Child + Mass of Pogo stick)a.
Then set that equal to -kx. But as you can see there are two unknowns.

If I'm neglecting any other principle that would help solve this problem. I'd be most gracious if you could tell me. Thanks
 
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  • #2
What two unknowns?

You know how much acceleration is produced by the upward restoring force, and what mass is being accelerated. Therefore, you know the upward restoring force. You also know the spring constant. Therefore, know the displacement from the equilibrium spring length.
 
  • #3
cepheid said:
What two unknowns?

You know how much acceleration is produced by the upward restoring force, and what mass is being accelerated. Therefore, you know the upward restoring force. You also know the spring constant. Therefore, know the displacement from the equilibrium spring length.

If I follow what you are saying. You want me to find the upward resulting force from the spring compression using the given mass of 28.86kg and multiply that my the acceleration given 4.802 m/s^2 and set that equal to the spring constant times the distance x.

F=(mass given)(acceleration given)
F=28.86kg*4.802m/s^2
F=138.58572N

138.58572N=-18016N/m*-deltax
deltax=0.007692369m

I did this, but the answer to my solution was wrong. Did I miss understand you? Am I missing something?
 
  • #4
Ok I solved it
 
  • #5
Prophet029 said:
F=(mass given)(acceleration given)
F=28.86kg*4.802m/s^2
F=138.58572N

138.58572N=-18016N/m*-deltax
deltax=0.007692369m

For those who are solving any problem like such in the future.

The original equation I used was correct but you have to factor in that you also have the force of gravity that acted on the spring as well. So the Total net force is:

Fnet=((mass given)*(acceleration given)) + ((mass given)*(gravity acceleration) which equals
kx form hooke's law F=-kx

equation form:
Fnet=([(mass given)*(acceleration given)] + [(mass given)*(gravity acceleration)])

Fnet=-kx (since compression in this case find positive values so neglect the negative)

([(mass given)*(acceleration given)] + [(mass given)*(gravity acceleration)])=kx

and solve for x
 
Last edited:

1. What is Hooke's Law?

Hooke's Law is a scientific principle that describes the relationship between the force applied to a spring and the resulting deformation or compression of the spring.

2. How is Hooke's Law used to find spring compression?

Hooke's Law can be used to calculate the amount of compression in a spring by using the equation F = kx, where F is the force applied, k is the spring constant, and x is the displacement or compression of the spring.

3. What is the significance of the spring constant in Hooke's Law?

The spring constant, also known as k, is a measure of the stiffness of a spring. It is unique to each spring and is used to calculate the amount of force required to produce a certain amount of compression in the spring.

4. Can Hooke's Law be applied to all types of springs?

Yes, Hooke's Law can be applied to all types of springs, as long as they follow the principle of linear elasticity. This means that the spring must return to its original shape and size after the force is removed.

5. What are some real-life applications of Hooke's Law and finding spring compression?

Hooke's Law and finding spring compression have many real-life applications, including designing and testing mechanical and electrical components, measuring the elasticity of materials, and developing devices such as shock absorbers, springs in mattresses, and more.

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