Vertical Mass Spring System | Analyzing work

[I_Try_Math]

Homework Statement

A massless spring with force k = 200 N/m constant hangs from the ceiling. A 2.0-kg block is attached to the free end of the spring and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?

Relevant Equations

F = kx

Is there a typo in this question? Supposing there was no friction, the block would fall until the force of the spring was equal to $mg = 2 * 9.8 = 19.6$, taking the upward y direction as positive. Since $F_{spring} = -200y$ and $19.6 = -200(-0.098)$, the block would fall 9.8 cm. It's not possible for the block to fall farther than that with friction opposing the motion?

 

[Orodruin]

The point when $F_{\rm spring} = -mg$ is not the turn around point. It is the point where the mass is no longer accelerating downwards, but instead starts to slow down. Your contradiction therefore rests on a false premise.

 

[I_Try_Math]

The point when $F_{\rm spring} = -mg$ is not the turn around point. It is the point where the mass is no longer accelerating downwards, but instead starts to slow down. Your contradiction therefore rests on a false premise.

Interesting, that makes sense. Thank you for pointing that out.

 

[Orodruin]

Can you perhaps think of a different approach? Have you tried using energy methods?

 

[I_Try_Math]

Can you perhaps think of a different approach? Have you tried using energy methods?

Well, my initial thoughts were to try to use $ME_{final} = KE_{final} + PE_{final} + W_{friction} = 0 - 0 + W_{friction}$? But if that's correct it doesn't seem to get me anywhere.

Edit: However if I reconsider energy if there was no friction maybe I can solve it.

 

[Orodruin]

Are you certain you have the right LHS? Also, how do you define your potential energies?

 

[I_Try_Math]

Are you certain you have the right LHS? Also, how do you define your potential energies?

I'm not really certain if the LHS is correct, but at least in the case without friction $ME_{initial} = KE_{initial} + PE_{initial}$. And since $ME_{initial} = 0, PE_{initial} = 0$ and $KE_{initial} = 0$. $ME_{final} = ME_{initial}= 0 = KE_{final} + PE_{final} = 0 + PE_{final}$ which implies $PE_{final} = 0?$

Potential energy, the way I tried to define it, decreases in the -y direction.

 

[Orodruin]

Potential energy, the way I tried to define it, decreases in the -y direction.

Does it? Are you trying to account only for gravitational potential energy? That is not the only potential energy at work here!

 

[I_Try_Math]

Does it? Are you trying to account only for gravitational potential energy? That is not the only potential energy at work here!

I'm trying to make sense of the net potential energy, but if my equations relating the initial and final mechanical energy are right, then I'm struggling to see how it could be anything other than zero. I guess what I'm saying is that in my mind the potential energies of spring and gravitation are "cancelling" out but that's incorrect?

 

[Orodruin]

I'm trying to make sense of the net potential energy, but if my equations relating the initial and final mechanical energy are right, then I'm struggling to see how it could be anything other than zero.

You have not yet explained why you would think they are zero. What are the contributions to the potential energy? What are their mathematical expressions?

 

[I_Try_Math]

You have not yet explained why you would think they are zero. What are the contributions to the potential energy? What are their mathematical expressions?

It's kind of counterintuitive, but I guess what I was trying to say was basically that the potential energies of gravitation and the spring would cancel out. I mean, in the case without friction, if $ME_{final} = 0 = ME_{initial}$ then it has to be that $PE_{final} = 0$ because $KE_{final}=0$. But that's incorrect?

 

[kuruman]

Perhaps you might consider using the work-energy theorem that says that the sum of all the works done on the block is equal to the change in kinetic energy. $$\Delta K=\sum_i W_i.$$You know that, from start to finish, the change in kinetic energy is zero, so add up all the works and set them equal to zero.

 

[kuruman]

. . . if $ME_{final} = 0 = ME_{initial}$ then . . .

That cannot happen. The final and initial mechanical energies cannot be equal because we are told that there is friction that eats up some of the mechanical energy as the block is descending.

 

[I_Try_Math]

That cannot happen. The final and initial mechanical energies cannot be equal because we are told that there is friction that eats up some of the mechanical energy as the block is descending.

Is it correct to say that in the case without friction, at the turn around point when the mass reaches it lowest position, the net potential energy is zero? I understand that it's maybe irrelevant to this case with friction but I'm still trying to see if I understand it correctly.

 

[PeroK]

Is it correct to say that in the case without friction, at the turn around point when the mass reaches it lowest position, the net potential energy is zero? I understand that it's maybe irrelevant to this case with friction but I'm still trying to see if I understand it correctly.

Gravitational potential energy of zero depends where you define the zero point. It's only the change in GPE that is relevant.

 

[I_Try_Math]

Gravitational potential energy of zero depends where you define the zero point. It's only the change in GPE that is relevant.

If I pick the zero of gravitational potential energy to be at the spring's equilibrium point, what would the net potential energy be at the lowest position of the mass if it was like this problem except without friction?

 

[PeroK]

If I pick the zero of gravitational potential energy to be at the spring's equilibrium point, what would the net potential energy be at the lowest position of the mass if it was like this problem except without friction?

That should be clear. If the KE is zero at both points, then all the GPE lost is converted to EPE.

 

[Orodruin]

I am still missing some equations that are actually applied to the problem in question rather than just describing the general idea. I believe that if you start writing down the relevant equations then things will tend to materialise.

 

[I_Try_Math]

I am still missing some equations that are actually applied to the problem in question rather than just describing the general idea. I believe that if you start writing down the relevant equations then things will tend to materialise.

Thankfully, I was finally able to work through to the correct solution using energy considerations, as you said.