Energy Conservation - Rock compression

[Rock00]

A 7.94 kg stone is resting on a spring. The spring is compressed 10.2 cm by the stone.

(a) Calculate the force constant of the spring.
(b) The stone is pushed down an additional 28.6 cm and released. How much potential energy is stored in the spring just before the stone is released?
(c) How high above this new (lowest) position will the stone rise?

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Answer:
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Diagram:


(rock on top)
______
(
)
( k ==> Spring
)
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(I have attached a better diagram of the above image)

Given
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m= 7.94 kg
spring compressed = .102m

I know the total change in energy is 0, so how do I start on the problem?

What formulas should I (can you please list it out)?

Any help would be appreciate it!

 

[andrewchang]

the first thing you will need to find is the spring constant- you can do this using the weight of the rock and the compression of the spring. the rest of the problems are solved with conservation of energy.

you need to show your work, or at least what you've tried.

 

[Rock00]

the first thing you will need to find is the spring constant- you can do this using the weight of the rock and the compression of the spring. the rest of the problems are solved with conservation of energy.
you need to show your work, or at least what you've tried.

The work I have provided is what I knew. I didn't know what to do.
By the way can you help me out with the Conservation of energy question. A physics guru by the name of Marlon says $$y = \frac{gt^2}{2}$$. How did he get that. Help would be appreciated.

 

[LeonhardEuler]

The work I have provided is what I knew. I didn't know what to do.
By the way can you help me out with the Conservation of energy question. A physics guru by the name of Marlon says $$y = \frac{gt^2}{2}$$. How did he get that. Help would be appreciated.

You probably have a formula in whatever textbook you are using that says that position, x, under constant acceleration is given by:
$$x=x_0+v_{0x}t+\frac{1}{2}at^2$$
In the case of an object accting under the influence of only gravity, this becomes
$$y=y_0+v_{0y}t+\frac{1}{2}gt^2$$
These equations are just obtained from integrating the acceleration twice:
$$\frac{d^2x}{dt^2}=a$$
$$\frac{dx}{dt}=at+C; C=v_{0x} \rightarrow \frac{dx}{dt}=v_{0x} + at$$
$$x=v_{0x}t + \frac{1}{2}at^2 +C_2; C_2=x_0 \rightarrow x=x_0 + v_{0x}t + \frac{1}{2}at^2$$

 

[Rock00]

Help

You probably have a formula in whatever textbook you are using that says that position, x, under constant acceleration is given by:
$$x=x_0+v_{0x}t+\frac{1}{2}at^2$$
In the case of an object accting under the influence of only gravity, this becomes
$$y=y_0+v_{0y}t+\frac{1}{2}gt^2$$
These equations are just obtained from integrating the acceleration twice:
$$\frac{d^2x}{dt^2}=a$$
$$\frac{dx}{dt}=at+C; C=v_{0x} \rightarrow \frac{dx}{dt}=v_{0x} + at$$
$$x=v_{0x}t + \frac{1}{2}at^2 +C_2; C_2=x_0 \rightarrow x=x_0 + v_{0x}t + \frac{1}{2}at^2$$

Can you give me advice as to how I would find Force constant, the potential energy and the displacement.

Thank you

 

[LeonhardEuler]

Oh, well that $$y = \frac{gt^2}{2}$$ really doesn't help with this question. To find the force constant of the spring balance the forces in the vertical direction. Acting down is the force of gravity, mg. Acting up is the spring force, kx.

As for the potential energy: if what you gave is the full description of the problem, then the question is ill-posed. You can take the point of zero gravitational potential energy to be anywhere you please and you'll get different answers depending on where you choose that to be. It is only changes in potential energy which are physically meaningful, not their absolute values. Are you sure they didn't tell you that the PE was taken to be zero at some stage? Anyway, the way to proceed is to add the PE from gravity, mgh, and the PE of the spring (1/2)kx².

To find the hieght to which the stone rises, apply conservation of energy. The rock had a certain amount of energy, (all potential) at the bottom position due to both gravity and the spring. Once it reaches its maximum height it is instantaneously at rest. This means its energy is once again all potential, but this time all due to gravity. Set the energies equal.