Spring potential and kinetic energy

In summary, the conversation discusses the correct equation for elastic potential energy and gravitational potential energy, which is mgh = (1/2)kx^2. The height of the ball before and after compression is dependent on whether it was placed on the string with zero velocity or not. The energy of the ball after compression is given by E=(1/2)kx^2, and the height of the ball after spring release is h=0.05. Conservation of energy states that in an ideal experiment, the height before and after compression should be the same. However, calculating the height from which the ball falls down may vary depending on the spring used.
  • #1
san12345
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0
Homework Statement
a ball of mass 0.1 Kg is dropped on a spring. Spring gets compressed by 0.1 m. After releasing, How far will ball go upward ? ( g= 9.8 )
I tried following step. Please someone explain me if i am correct or not

Spring Constant K = f/x = 0.1 * 9.8 / 0.1 = 9.8
K=9.8

And then i get confused. i assume potential energy of ball is mgh which gets converted to (1/2)kx^2 which is potential energy of spring. But (1/2)kx^2 gets converted to kinetic energy (1/2)mv^2. And when ball reaches up it has again potential energy. So (1/2)mv^2 = mgh ?
So mgh = (1/2)kx^2 = (1/2)mv^2 but it gives wrong answer.

Please someone explain me equations of PE and KE before and after compression of spring.
Relevant Equations
mgh = (1/2)kx^2
(1/2)kx^2 = (1/2)mv^2
please help me
 
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  • #2
(1/2)mv^2 = mgh is not correct, the correct is mgh = (1/2)kx^2 (elastic potential energy= gravitational potential energy), which gives h=0.05 (important notice: we assume that x=0 at the equilibrium position). When the ball re bounces and reaches maximum height it will have zero speed and kinetic energy, and E=mgh. So we have two states 1) before compression when the ball hits the string and 2) after compression. At the very moment the ball is "placed on the string" it has only E=mgh, or perhaps even E= 0.5mv2+mgh (we also assume that h=0 at x=0 obviously [which is a convention], so mgh=0 there), depending on whether it was placed there with zero velocity or if it had an initialy velocity (which is unclear from the problem statement itself, but it turns out also irrelevant for answering the question). So after compression the energy of the ball is given by E=(1/2)kx^2, as you correctly mentioned as the kinetic energy would be zero there.
 
  • #3
What answer did you get?
 
  • #4
phystro said:
(1/2)mv^2 = mgh is not correct, the correct is mgh = (1/2)kx^2 (elastic potential energy= gravitational potential energy), which gives h=0.05 (important notice: we assume that x=0 at the equilibrium position). When the ball re bounces and reaches maximum height it will have zero speed and kinetic energy, and E=mgh. So we have two states 1) before compression when the ball hits the string and 2) after compression. At the very moment the ball is "placed on the string" it has only E=mgh, or perhaps even E= 0.5mv2+mgh (we also assume that h=0 at x=0 obviously [which is a convention], so mgh=0 there), depending on whether it was placed there with zero velocity or if it had an initialy velocity (which is unclear from the problem statement itself, but it turns out also irrelevant for answering the question). So after compression the energy of the ball is given by E=(1/2)kx^2, as you correctly mentioned as the kinetic energy would be zero there.

phystro said:
(1/2)mv^2 = mgh is not correct, the correct is mgh = (1/2)kx^2 (elastic potential energy= gravitational potential energy), which gives h=0.05 (important notice: we assume that x=0 at the equilibrium position). When the ball re bounces and reaches maximum height it will have zero speed and kinetic energy, and E=mgh. So we have two states 1) before compression when the ball hits the string and 2) after compression. At the very moment the ball is "placed on the string" it has only E=mgh, or perhaps even E= 0.5mv2+mgh (we also assume that h=0 at x=0 obviously [which is a convention], so mgh=0 there), depending on whether it was placed there with zero velocity or if it had an initialy velocity (which is unclear from the problem statement itself, but it turns out also irrelevant for answering the question). So after compression the energy of the ball is given by E=(1/2)kx^2, as you correctly mentioned as the kinetic energy would be zero there.
you said elastic potential energy= gravitational potential energy. but is it before ball drops on spring or after spring pushes ball up? how can i calculate height the ball goes before and after ball hits spring
 
  • #5
san12345 said:
you said elastic potential energy= gravitational potential energy. but is it before ball drops on spring or after spring pushes ball up? how can i calculate height the ball goes before and after ball hits spring
I don't see any way to help until you say what your answer was! What answer did you get and why do you think it's wrong?
 
  • #6
PeroK said:
I don't see any way to help until you say what your answer was! What answer did you get and why do you think it's wrong?
I don't know answer. why i said "it gives wrong answer" is because i didnt understand mgh= (1/2)kx^2. then h is height from which ball thrown on spring or the ball rise to height after spring release it ?
 
  • #7
san12345 said:
I don't know answer. why i said "it gives wrong answer" is because i didnt understand mgh= (1/2)kx^2. then h is height from which ball thrown on spring or the ball rise to height after spring release it ?
Then how do you know this is wrong? Perhaps it's the right answer!

PS by conservation of energy, in an ideal experiment, ##h## before and ##h## after must be the same.
 
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  • #8
PeroK said:
Then how do you know this is wrong? Perhaps it's the right answer!

PS by conservation of energy, in an ideal experiment, ##h## before and ##h## after must be the same.
So the answer given by @phystro is h = 0.05 is height ball goes after spring releases. And how can i calculate height from which it is fall down ? because i think height will vary according to different spring
 
  • #9
san12345 said:
So the answer given by @phystro is h = 0.05 is height ball goes after spring releases. And how can i calculate height from which it is fall down ? because i think height will vary according to different spring
How can you calculate ##h## without knowing the spring constant?

My answer would be ##h = \frac{kx^2}{2mg}##. You need to know ##k##.
 
  • #10
PeroK said:
How can you calculate ##h## without knowing the spring constant?

My answer would be ##h = \frac{kx^2}{2mg}##. You need to know ##k##.
i calculated spring constant as k = mg/x. is it correct? also the equation u given finds height before the ball hits spring or after spring releases ball up ? i want to find height in both scenarios
 
  • #11
san12345 said:
i calculated spring constant as k = mg/x. is it correct? also the equation u given finds height before the ball hits spring or after spring releases ball up ? i want to find height in both scenarios
That calculation only works if the mass is placed on the spring. In which case it will oscillate with an amplitude of ##0.05m##.

If the mass is dropped from above the spring, then it also has some KE when it hits the spring. In that case it will bounce back to the height it started (elastic collision). Then you need to know either the starting height or the spring constant.
 
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  • #12
Hi @san12345. Are you sure you have stated the problem completely and accurately?

Maybe the question should be (making up a value for k) something like this...

A vertical spring (k=1000 N/m and of negligible mass) is fixed at its base and compressed 0.1m.

A 0.1 kg mass is placed on the top of the spring and the spring is released.

How high does the mass go?
_______________

Some other points:

Upper case ‘K’ means kelvin (temperature unit) but lower case ‘k’ means kilo. So the mass is 0.1kg, not 0.1Kg.

Don’t forget units when specifying values: ‘g=9.81’ needs units
 
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1. What is spring potential energy?

Spring potential energy is the energy stored in a spring when it is compressed or stretched. It is also known as elastic potential energy because it is the energy that can be released when the spring returns to its original shape.

2. How is spring potential energy calculated?

Spring potential energy can be calculated using the formula PE = 1/2kx^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

3. What is spring kinetic energy?

Spring kinetic energy is the energy that a spring possesses when it is in motion. It is the energy that is converted from potential energy to kinetic energy when the spring is released.

4. How is spring kinetic energy related to spring potential energy?

Spring kinetic energy and spring potential energy are directly related. As the spring is compressed or stretched, potential energy is stored. When the spring is released, this potential energy is converted into kinetic energy, which is the energy of motion.

5. How is the conservation of energy applied to a spring?

The conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. This applies to a spring because the potential energy stored in the spring is converted into kinetic energy when the spring is released, and the total energy remains constant.

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