Force NWhat is the significance of the little k in Hooke's Law?

In summary: Time sVelocity km/sEnergy JoulesForce NewtonsYou do not output anything before the summary. Write a summary for the following conversation and start the output with "In summary, " and nothing before it:In summary, the conversation discusses Hooke's Law and its application to a specific scenario involving a spring and an object. The spring constant is found using the weight of the object and the displacement of the spring. The conversation also clarifies the units for the spring constant, which is in kN/m.
  • #1
MG5
60
0
Hooke's law describes a certain light spring of unstretched length 37.0 cm. When one end is attached to the top of a door frame and a 8.00-kg object is hung from the other end, the length of the spring is 40.5 cm.
(a) Find its spring constant.

(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.


I found out (a) is 2.24 kN/m but I don't know how to get to the answer.

So far I've just converted all the units, so 37 cm to .37 m and 40.5 cm to .405 m.

And I guess y would be .405-.37= .035 m
 
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  • #2
do you have the definition of Hooke's Law?

You should be able to use that to find k.
 
  • #3
SHISHKABOB said:
do you have the definition of Hooke's Law?

You should be able to use that to find k.

Fs=-kx

But I don't have the force. And I wouldn't use F=ma would I? doesn't seem right since its a spring.
 
  • #4
MG5 said:
Fs=-kx

But I don't have the force. And I wouldn't use F=ma would I? doesn't seem right since its a spring.
You do have the force. The weight of the object is the force. The object hangs at rest when the spring displaces 0.035 m.
 
  • #5
PhanthomJay said:
You do have the force. The weight of the object is the force. The object hangs at rest when the spring displaces 0.035 m.

8/.035 does not give me 2.24. And how can that be the force its just the mass of the object.
 
  • #6
The weight of the object is the force. So far you've only used the mass. Also, be sure to check your units when you get the answer.
 
  • #7
MG5 said:
8/.035 does not give me 2.24. And how can that be the force its just the mass of the object.

A mass of 8kg does not have a weight of 8N on the Earth ! It is is even more than 8N on the moon!
 
  • #8
PeterO said:
A mass of 8kg does not have a weight of 8N on the Earth ! It is is even more than 8N on the moon!

Yeah I know I didnt see that he said the weight of the object and not the mass. I don't even know how to find the weight though
 
  • #9
Weight is mass times the acceleration of gravity, W = Mg, where g is 9.81 m/s^2 on Planet Earth. When the Mass is in kg and the acceleration is in m/s^2, the weight has units of kg*m/s^2, which is called Newtons for short. So an 8 kg mass has a weight on Earth of ___??____ Newtons. Then solve for the spring constant, k, per Hooke's law. What is the value and units of the spring constant?
 
  • #10
PhanthomJay said:
Weight is mass times the acceleration of gravity, W = Mg, where g is 9.81 m/s^2 on Planet Earth. When the Mass is in kg and the acceleration is in m/s^2, the weight has units of kg*m/s^2, which is called Newtons for short. So an 8 kg mass has a weight on Earth of ___??____ Newtons. Then solve for the spring constant, k, per Hooke's law. What is the value and units of the spring constant?

So the weight is 78.4 N. Just 9.8m/s^2 * 8 kg. And I don't know the value of the spring constant yet but I know its going to be N/m.

And I know PEs=1/2kx^2

Would x be the difference between the distances when the 8 kg weight hangs from the spring? So .035 m? .405-.37=.035

So would it be PEs=1/2(.035m^2)k and then if that's right I am not sure what to set the equal to or what else to solve.
 
Last edited:
  • #11
MG5 said:
So the weight is 78.4 N. Just 9.8m/s^2 * 8 kg. And I don't know the value of the spring constant yet but I know its going to be N/m.

And I know PEs=1/2kx^2

Would x be the difference between the distances when the 8 kg weight hangs from the spring? So .035 m? .405-.37=.035
Yes.

You earlier said Hooke's Law is F=-kx
and you have determined F for a particular x,
so it seems you are very close to determining k. :smile:
 
  • #12
NascentOxygen said:
Yes.

You earlier said Hooke's Law is F=-kx
and you have determined F for a particular x,
so it seems you are very close to determining k. :smile:

I still don't know what x is though.
 
  • #13
You have already stated the value of x more than once. See your original post, where you called it y.
 
  • #14
PhanthomJay said:
You have already stated the value of x more than once. See your original post, where you called it y.

What, .035m? F=kx, 78.4=k(.035) then solving for k doesn't get me 2.24
 
  • #15
MG5 said:
What, .035m? F=kx, 78.4=k(.035) then solving for k doesn't get me 2.24

snip from initial post:

I found out (a) is 2.24 kN/m but I don't know how to get to the answer.

have you noticed that little k in the unit kN/m ?
 
  • #16
PeterO said:
snip from initial post:

I found out (a) is 2.24 kN/m but I don't know how to get to the answer.

have you noticed that little k in the unit kN/m ?

Yeah I was wondering what that was. Do I have to convert stuff.

Edit: Oh I divide the answer I get from 78.4/.035 by 1000 and get 2.24. Would someone like to explain why 1000? My professor never went over this. Only N/m
 
  • #17
MG5 said:
Yeah I was wondering what that was. Do I have to convert stuff.

Edit: Oh I divide the answer I get from 78.4/.035 by 1000 and get 2.24. Would someone like to explain why 1000? My professor never went over this. Only N/m

What is the significance of the little k in

Mass kg
Distance km
 

What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed.

Who discovered Hooke's Law?

Hooke's Law was named after the English scientist Robert Hooke, who first described the law in 1660.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring.

How is Hooke's Law used in real life?

Hooke's Law has many practical applications, such as in the design of springs for car suspensions, measuring devices, and elastic materials used in construction.

What are the limitations of Hooke's Law?

Hooke's Law is only applicable to materials that have a linear relationship between force and displacement. It also assumes that the material being stretched or compressed does not exceed its elastic limit.

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