The value of the spring constant k that I calculate seems too high

  • #1
dannolul
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Homework Statement
A ball of mass 73kg starts at the top of a 52m high hill with a speed of 20.0 m/s. It rolls down the hill onto a spring and compresses it to a maximum of 46.5cm. Find the spring constant of the spring.
Relevant Equations
Et1=Et2
ek1+eg1+ee1=ek2+eg2+ee2
ek1+eg1=Ee2

Ek= kinetic energy (1/2mv^2)
eg=Grabitational energy (mgh)
Ee=elastic energy (1/2kx^2)
I expanded ET1=ET2 to get

(Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom)



Rearanged i got



k = (mv^2+2mgh)/x^2



so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2

=479137.945N/m
 
Last edited:
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  • #2
Please show the details of your calculation. Without them, we cannot find where and if you went wrong. Also, when you quote numbers you must attach units to them. 479000 is meaningless all by itself.
 
  • #3
kuruman said:
Please show the details of your calculation. Without them, we cannot find where and if you went wrong. Also, when you quote numbers you must attach units to them. 479000 is meaningless all by itself.
Oh Sorry, was on my phone and rushing.

I expanded ET1=ET2 to get
(Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom)

Rearanged i got

k = (mv^2+2mgh)/x^2

so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2
=479137.945N/m
 
Last edited:
  • #4
dannolul said:
Oh Sorry, was on my phone and rushing.

I expanded ET1=ET2 to get
(Total energy at top) 1/2mv^2+mgh = 1/2kx^2 (Total energy at bottom)

Rearanged i got

k = (mv^2+2mgh)/x^2

so [(73)(20)^2+2(73)(9.8)(52)]/0.465^2
=479137.945N*m
You mean 479000 N/m. That’s what I got. Why do you think it’s too high? You have a 73 kg mass starting at the top of a 50 m hill initially at 45 miles per hour. When it starts compressing the spring, it’s moving at about 84 miles per hour. It must be quite a spring to stop that mass over a distance of about 2 feet.
 
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  • #5
kuruman said:
You mean 479000 N/m. That’s what I got. Why do you think it’s too high? You have a 73 kg mass starting at the top of a 50 m hill initially at 45 miles per hour. When it starts compressing the spring, it’s moving at about 84 miles per hour. It must be quite a spring to stop that mass over a distance of about 2 feet.
In class we havnt really talked about high k constant and only worked with small k constants. So it seemed a little skeptical as I didnt think it should be that high. Appreciate the help.
 
  • #6
Other than the unusual input numbers, I am bothered by the solution that conserves energy without considering the rotation of the ball that is clearly said to "roll down the hill." Have you studied rotational motion? If so, you can recalculate the speed of the center of mass of the wheel taking the rotation of the wheel into consideration.

However, in that case one has to consider how the kinetic energy of the wheel is converted to elastic potential energy in the spring without energy dissipation by friction as the ball rolls (?) an additional 46.5 cm before it comes to rest.
 
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  • #7
kuruman said:
Other than the unusual input numbers, I am bothered by the solution that conserves energy without considering the rotation of the ball that is clearly said to "roll down the hill." Have you studied rotational motion? If so, you can recalculate the speed of the center of mass of the wheel taking the rotation of the wheel into consideration.

However, in that case one has to consider how the kinetic energy of the wheel is converted to elastic potential energy in the spring without energy dissipation by friction as the ball rolls (?) an additional 46.5 cm before it comes to rest.
I suspect it was just unintentional wording. No radius for the ball either.

EDIT: never mind that is independent of the radius, I still suspect poor wording.

What would stop it rotation at the bottom; the friction with the ground? if so, all of its rotational kinetic energy would be lost to heat regardless, no? Or if there was no friction at the bottom (on either the ground or the spring) it would just be spinning freely at whatever angular velocity it had as it compressed the spring.
 
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  • #8
erobz said:
What would stop it rotation at the bottom; the friction with the ground? if so, all of its rotational kinetic energy would be lost to heat regardless, no? Or if there was no friction at the bottom (on either the ground or the spring) it would just be spinning freely at whatever angular velocity it had as it compressed the spring.
Exactly. One has to assume that there is no friction at the bottom. In that case only the translational kinetic energy will be converted to elastic energy which is results in the solution already obtained.
 
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  • #9
kuruman said:
Other than the unusual input numbers, I am bothered by the solution that conserves energy without considering the rotation of the ball that is clearly said to "roll down the hill."
What bothers me more is that mechanical energy would not be anywhere near conserved if you actually did what's described in this scenario.
 
  • #10
erobz said:
if so, all of its rotational kinetic energy would be lost to heat regardless, no?
You want to be careful with statements like this and the misconceptions they generate. Heat is a transfer of energy due to a temperature difference. The correct statement is that the rotational kinetic energy (or indeed any form of mechanical energy) is converted to internal energy.

This distinction becomes important when you study the First Law of Thermodynamics.
 
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  • #11
Mister T said:
What bothers me more is that mechanical energy would not be anywhere near conserved if you actually did what's described in this scenario.
It's not a good problem. I give it a C-.
 
  • #12
kuruman said:
One has to assume that there is no friction at the bottom.
The normal force from the spring will (soon) be much greater than that from the ground. Even a small frictional coefficient against the spring will lift it off the ground completely.
 

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