Derive the equation for kinetic energy and gravitional energ

AI Thread Summary
The discussion revolves around calculating kinetic energy, elastic potential energy, gravitational potential energy, and total energy for a wooden toy mouse attached to a spring at various positions. At the equilibrium point, where the spring is unstretched (x=0), the elastic potential energy is zero, but the kinetic energy is derived as KE=1/2m*sqrt(k/m)*A. Participants debate whether the elastic potential energy can truly be zero at this point, emphasizing that while the forces balance, the spring still stores energy. The conversation highlights the importance of understanding the definitions of potential and kinetic energy in the context of the system's dynamics. Clarifications are sought regarding the role of gravitational force and tension in the spring at equilibrium.
Derek1997
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Homework Statement


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 gravitational 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)

Homework Equations


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

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 canceled the term and derived the equation 1/2 mw^2. Could you please tell me if I am right?
 

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Derek1997 said:
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 canceled the term and derived the equation 1/2 mw^2.
That is not at all clear. Please clarify which part of the question you are discussing and post your detailed working.
 
Hi thanks for reply,
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?
 
Derek1997 said:
the equation for potential energy as PE= 1/2kx^2
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?
 
well x=0, so by that it would make the whole equation to 0?
 
Derek1997 said:
well x=0, so by that it would make the whole equation to 0?
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?
 
i think so. well how can it have it when it's 0?
 
Derek1997 said:
i think so. well how can it have it when it's 0?
Your equation says it is zero. I am challenging your equation.
What is meant by "equilibrium point" in this context?
 
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
Derek1997 said:
where it remained same, no upward or downward motion,
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
gravitional potential and normal force? but don't you just need to state the equation?
 
  • #12
Derek1997 said:
gravitional potential and normal force? but don't you just need to state the equation?
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
yea true sorry. mouse is not on the ground.
 
  • #14
Derek1997 said:
yea true sorry. mouse is not on the ground.
So, ik gravition is the force but what does it do with equation?
 
  • #15
Derek1997 said:
So, ik gravition is the force but what does it do with equation?
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
tension on the spring? balances it.
 
  • #17
Derek1997 said:
tension on the spring? balances it.
Right. If there is tension in the spring, how can it not be storing any elastic energy?
 
  • #18
yea but they don't balance out by gravitational energy?
 
  • #19
btw am i right with the kinetic energy one??
 
  • #20
Derek1997 said:
yea but they don't balance out by gravitational energy?
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.
 
  • #21
haruspex said:
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.
so cancells out?
 
  • #22
Derek1997 said:
so cancells out?
Cancels out in terms of the motion of the mouse, but not in terms of the elastic PE (which is what the question asks for).
 
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