# Spring potential and kinetic energy

• san12345

#### san12345

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

(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.

(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.

(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

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?

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 ?

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.

• Steve4Physics
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

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##.

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

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.

• Lnewqban and Steve4Physics
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

• PeroK