Spring potential energy and mass

In summary, the mass of the spring and the mass of the Earth-ball-spring system changes by 6.597×10-6 kilograms during the time interval from t=0 to the instant the ball reaches its maximum height.
  • #1
kolua
69
3

Homework Statement


One end of a vertical spring of spring constant k = 1900 N/m is attached to the floor. You compress the spring so that it is 2.50 m shorter than its relaxed length, place a 1.00-kg ball on top of the free end, and then release the system att = 0. (All values are measured in the Earth reference frame.)

A. By how much does the mass of the spring change during the time interval from t = 0 to the instant the ball leaves the spring?

B. By how much does the mass of the Earth-ball-spring system change during the time interval from t = 0to the instant the ball reaches its maximum height?

Homework Equations


ΔE=Δmc2
Us=1/2 kx2

The Attempt at a Solution


ΔE=ΔU=1900/2×2.52=5937.5
Δm=5937.5/9/188=6.597×10-6
is this correct for question A?
I'm not sure if the mass changes because the instant the ball leaves the potential energy is converted to the ball's kinetic energy. And there should be no change in mass?

For question B, how should I approach?
 
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  • #2
kolua said:
ΔE=ΔU=1900/2×2.52=5937.5
Δm=5937.5/9/188=6.597×10-6
is this correct for question A?
That looks right for a).
For b), what energy has entered or left the Earth-spring-mass system?
 
Last edited:
  • #3
kolua said:
I'm not sure if the mass changes because the instant the ball leaves the potential energy is converted to the ball's kinetic energy.

Well, it's not an instantaneous process. As the spring relaxes the ball picks up speed, so the process takes as long as it takes for the spring to go from it's compressed state to its relaxed state. And during the process the ball rises, so there's also an increase in the gravitational potential energy of the ball-Earth system.

And there should be no change in mass?

Change in mass of the spring, or change in mass of the whole system?
 
  • #4
haruspex said:
That looks right for a).
For b), what energy has entered or left the Earth-spring-mass system?
potential energy and kinetic energy?
 
  • #5
kolua said:
potential energy and kinetic energy?
Those are both internal to the Earth-spring-mass system.
 
  • #6
haruspex said:
Those are both internal to the Earth-spring-mass system.
So should I count the mass of the ball and the Earth intot the calculation for this one?
 
  • #7
What haruspex is trying to get you to do is determine which of these energies is leaving or exiting the system. Note that the choice of system is arbitrary. It's different in Parts A and B of this problem.
 
  • #8
Mister T said:
What haruspex is trying to get you to do is determine which of these energies is leaving or exiting the system. Note that the choice of system is arbitrary. It's different in Parts A and B of this problem.
For question A, since there is only the spring in the system, there is only the change in the spring potential energy, which is U=1/2 kx2 = E=mc2, is this correct?
For question B, there are potential and kinetic energy for this system, so U=1/2 kx2=E-K? is this right?
 
  • #9
Best to think of it in terms of states of the system. The initial state has energy ##E_i## and the final state has energy ##E_f##. In this way you can account for kinetic energy and both types of potential energy. Something you're not doing now is including both types of potential energy.
 
  • #10
Mister T said:
Best to think of it in terms of states of the system. The initial state has energy ##E_i## and the final state has energy ##E_f##. In this way you can account for kinetic energy and both types of potential energy. Something you're not doing now is including both types of potential energy.
Uspring=1/2 kx2=E-K-Ugravity?
 
  • #11
Ok, but you're not solving for ##U_{spring}##.
 
  • #12
Mister T said:
Ok, but you're not solving for ##U_{spring}##.
Δmc2=K+Ugravity?
 
  • #13
I think you should explain briefly what you're doing to get these expressions. In particular, go back and read the statement of the problem, then read Post #9 and follow the scheme outlined there.
.
 

What is spring potential energy?

Spring potential energy is the energy stored in a spring when it is compressed or stretched from its equilibrium position. It is a type of potential energy that can be converted into other forms, such as kinetic energy, when the spring is released.

How is spring potential energy calculated?

The formula for calculating spring potential energy is E = 1/2kx², where E is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position. This formula assumes that the spring is ideal and obeys Hooke's Law.

What is the relationship between mass and spring potential energy?

The amount of spring potential energy stored in a spring is directly proportional to the mass attached to the spring. This means that as the mass increases, so does the potential energy stored in the spring. This relationship is described by the formula E = 1/2kx², where x is the displacement and k is the spring constant.

Can mass affect the spring constant?

No, mass does not affect the spring constant. The spring constant is a measure of the stiffness of the spring and remains constant regardless of the mass attached to the spring. However, the displacement of the spring will change with the mass, which will in turn affect the amount of potential energy stored in the spring.

How does the spring constant affect the potential energy of a spring?

The spring constant directly affects the potential energy of a spring. A higher spring constant means that the spring is stiffer and requires more force to compress or stretch it. This results in a higher potential energy stored in the spring for the same displacement. Conversely, a lower spring constant means that the spring is less stiff and will have a lower potential energy for the same displacement.

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