Help needed in this problem involving a spring and energy balances

In summary, the problem involves analyzing a spring system where energy balances must be applied to understand the dynamics of the spring's motion. Key concepts include the potential energy stored in the spring, kinetic energy of the moving mass, and the principles of conservation of energy to solve for various parameters such as displacement, velocity, and forces acting on the system. Assistance is sought in applying these principles effectively to arrive at a solution.
  • #1
hello478
165
14
Homework Statement
need help in part c of the questions
see question attached below
i dont understand how gpe can be equal to strain energy
in fig 5.2
it is in equillibrium so it wont have gpe or strain energy
but as it moves down it gains gpe and strain energy
then the total energy of system should be zero
thinking about it another way
will the gpe it gain be negative??
so this way the total=gpe+strain energy
and because initial = 0 the final would also be zero
so 0=1/2kx^2 - 0.024
and then 0.025=1/2kx^2
am i right??please correct me
what if in this question kinetic energy is also involved???
Relevant Equations
energy equations... mgh, 1/2kx^2
1709931702774.png
1709931730017.png

i do know how to do the working but i dont understand the concept stated above...
 
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  • #2
According to the homework help guidelines, you need to show some effort. It's not good enough just to say you don't know.
 
  • #3
hello478 said:
it is in equillibrium so it wont have gpe or strain energy
but as it moves down it gains gpe and strain energy
GPE is relative to some defined zero. The text only refers to changes in GPE, so it does not matter where you set the zero.
As it moves down it loses GPE, so if you define it as zero at the relaxed position then it is going negative, which is fine.
But you are wrong to call 5.2 the equilibrium position; it is the relaxed position. 5.1 is equilibrium.

The question asks for the "change" in GPE. I would take that as a signed value, so negative.
Please explain how you arrived at 0.024 (or is it 0.025?) and state the units.
 
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  • #4
PeroK said:
According to the homework help guidelines, you need to show some effort. It's not good enough just to say you don't know.
please read the whole question, im sorry if you didnt get it, i did try to attempt and solve it above... but i solved it in the homework statement part, sorry about that, please read it from there
 
  • #5
haruspex said:
GPE is relative to some defined zero. The text only refers to changes in GPE, so it does not matter where you set the zero.
As it moves down it loses GPE, so if you define it as zero at the relaxed position then it is going negative, which is fine.
But you are wrong to call 5.2 the equilibrium position; it is the relaxed position. 5.1 is equilibrium.

The question asks for the "change" in GPE. I would take that as a signed value, so negative.
Please explain how you arrived at 0.024 (or is it 0.025?) and state the units.
mgh = 0.094*9.81*(2.6/1000) = 2.39 * 10^-3 J
but for part c... we are not looking at figure 5.1 so when it is stopped by a hand/ force so cant we say its in equilibrium?
 
  • #6
hello478 said:
in fig 5.2
it is in equillibrium so it wont have gpe or strain energy
but as it moves down it gains gpe and strain energy
What is it for you, the spring or the mass?

If it refers to the spring, "as it moves down it gains gpe" means (to me at least) that the spring gains or accumulates elastic potential energy (not gpe) as it gets stretched by gravity.

If it refers to the mass, "as it moves down it gains gpe" would be incorrect because the mass loses gravitational potential energy as it falls down.
 
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  • #7
hello478 said:
mgh = 0.094*9.81*(2.6/1000) = 2.39 * 10^-3 J
1000?
hello478 said:
but for part c... we are not looking at figure 5.1 so when it is stopped by a hand/ force so cant we say its in equilibrium?
I suppose you could argue that while it is being held it is in equilibrium, but then it is released, so it depends exactly what point in time you mean. Usually, if you describe a spring+mass system as being in equilibrium you mean without any forces on the mass except the spring and gravity.
Besides, you wrote:
"it is in equillibrium so it wont have gpe or strain energy"
If you meant while it is being held, the GPE and strain energy could be anything at all.
In position 5.2, it will indeed have no strain energy, but that is because the spring is relaxed, not because of any equilibrium.
 
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  • #8
haruspex said:
1000?

I suppose you could argue that while it is being held it is in equilibrium, but then it is released, so it depends exactly what point in time you mean. Usually, if you describe a spring+mass system as being in equilibrium you mean without any forces on the mass except the spring and gravity.
Besides, you wrote:
"it is in equillibrium so it wont have gpe or strain energy"
If you meant while it is being held, the GPE and strain energy could be anything at all.
In position 5.2, it will indeed have no strain energy, but that is because the spring is relaxed, not because of any equilibrium.
thank you soo much, makes a little sense now :)
 
  • #9
Lnewqban said:
What is it for you, the spring or the mass?

If it refers to the spring, "as it moves down it gains gpe" means (to me at least) that the spring gains or accumulates elastic potential energy (not gpe) as it gets stretched by gravity.

If it refers to the mass, "as it moves down it gains gpe" would be incorrect because the mass loses gravitational potential energy as it falls down.
yeah ill keep both views in mind
but the answer with both would be the same??
 

FAQ: Help needed in this problem involving a spring and energy balances

How do you calculate the potential energy stored in a spring?

The potential energy stored in a spring is calculated using Hooke's Law, which is given by the formula \( PE = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

What is the spring constant, and how do you determine it?

The spring constant, denoted as \( k \), is a measure of the stiffness of the spring. It can be determined experimentally by applying a known force to the spring and measuring the displacement. The spring constant is then calculated using the formula \( k = \frac{F}{x} \), where \( F \) is the applied force and \( x \) is the displacement.

How does energy conservation apply to a system involving a spring?

In a system involving a spring, energy conservation means that the total mechanical energy (potential plus kinetic) remains constant if there are no non-conservative forces (like friction) acting on the system. The potential energy stored in the spring can be converted to kinetic energy and vice versa, but the total energy remains the same.

What is the difference between elastic potential energy and kinetic energy in the context of a spring?

Elastic potential energy is the energy stored in the spring when it is compressed or stretched from its equilibrium position. It is given by \( PE = \frac{1}{2} k x^2 \). Kinetic energy, on the other hand, is the energy of motion and is given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In a spring system, energy oscillates between these two forms.

How do you solve problems involving a mass-spring system undergoing simple harmonic motion?

To solve problems involving a mass-spring system undergoing simple harmonic motion, you need to understand the relationship between displacement, velocity, and acceleration in the system. The motion can be described by the differential equation \( m\frac{d^2x}{dt^2} + kx = 0 \). The solution to this equation gives the position \( x(t) \) as a function of time, which typically takes the form \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency \( (\omega = \sqrt{\frac{k}{m}}) \), and \( \phi \) is the phase constant determined by initial conditions.

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