# Derive the equation for kinetic energy and gravitional energ

1. Oct 15, 2016

### Derek1997

1. The problem statement, all variables and given/known data
A wooden toy mouse of mass (m) is attached to a spring with constant (k) and suspended vertically as shown below. The toy is released at the point the spring is unstretched at position x = +A, passes through equilibrium at x = 0 and the spring’s maximum extension occurs at x = -A. If we ignore air friction and assume an ideal spring calculate expressions for the kinetic energy Ek, the elastic potential energy E elast, the gravitional potential energy E pot,and the total energy of the system Etotal, at (i) the release point and (ii) the equilibrium point. Take the gravitational potential energy to be zero at the equilibrium point y=0 and note that for a mass on a spring the maximum velocity vm=wA where w= sqrt (k/m)

2. Relevant equations
pe=1/2kx^2
e=mgh
X= acoswt
K=mw^2

3. The attempt at a solution
For equilibrium, I wrote for kinetic expression, since equation is 1/2 kx^2 i substitued the equation k mw^2, and since x was 0 and i cancelled the term and derived the equation 1/2 mw^2. Could you please tell me if im right?

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2. Oct 15, 2016

### haruspex

That is not at all clear. Please clarify which part of the question you are discussing and post your detailed working.

3. Oct 16, 2016

### Derek1997

so in regards to equilibrium point one, where x=o what i did was, i derived the equation for potential energy as PE= 1/2kx^2 where x=0 thus the whole equation would be zero and we don't need to worry about it.
So i used kinetic energy which yeilded me KE=1/2 mv^2, substituing, v as sqrt (k/m), thus:
KE= 1/2m*sqrt(k/m)*A
thus the total energy at equilibrium point would only depend on kinetic energy, which is E tot= KE which is the equation above.
am i following correct method?

4. Oct 16, 2016

### haruspex

It asks specifically for the elastic potential energy first, so I assume that is what you mean here.
It specifies x as being measured from the equilibrium point. Is the elastic PE really zero there?

5. Oct 16, 2016

### Derek1997

well x=0, so by that it would make the whole equation to 0?

6. Oct 16, 2016

### haruspex

It makes your equation 0, but that is not what I asked. What is happening to the spring at the equilibrium point? Does it really have no elastic PE?

7. Oct 16, 2016

### Derek1997

i think so. well how can it have it when it's 0?

8. Oct 16, 2016

### haruspex

What is meant by "equilibrium point" in this context?

9. Oct 16, 2016

### Derek1997

where it remained same, no upward or downward motion, could you please check if i have a right answers? if it is i can send you the second part.

10. Oct 16, 2016

### haruspex

Where it can remain static, yes... it is not necessarily static there ... and what will be the forces acting on it at that point?

11. Oct 16, 2016

### Derek1997

gravitional potential and normal force? but don't you just need to state the equation?

12. Oct 16, 2016

### haruspex

Gravitation, yes, but not potential - a potential is not a force.
What do you mean by normal force here? The toy mouse is not on the ground.

13. Oct 16, 2016

### Derek1997

yea true sorry. mouse is not on the ground.

14. Oct 16, 2016

### Derek1997

So, ik gravition is the force but what does it do with equation?

15. Oct 16, 2016

### haruspex

You claim that the elastic PE is zero at the equilibrium point. I am trying to prove to you that it is not, so your equation is wrong.
If gravity is acting, how come it is an equilibrium point? What is balancing the gravity?

16. Oct 16, 2016

### Derek1997

tension on the spring? balances it.

17. Oct 16, 2016

### haruspex

Right. If there is tension in the spring, how can it not be storing any elastic energy?

18. Oct 16, 2016

### Derek1997

yea but they don't balance out by gravitional energy?

19. Oct 16, 2016

### Derek1997

btw am i right with the kinetic energy one??

20. Oct 16, 2016

### haruspex

Sure, but the question asks specifically for the elastic PE at equilibrium. It is equal and opposite to the gravitational PE (measured from the initial position), but it is not zero.