# Energy Principle Problems

1. Oct 21, 2014

### The Wanderer

First off I just want someone to check and see if I got the right answer because I have no way of telling if it is the right answer or not. I am pretty sure I have the right answer but I have no way of checking.

1. The problem statement, all variables and given/known data

A spring with a stiffness ks and relaxed length L0 stands vertically on a table. You hold a mass M barely touching the top of the spring.

a) You very slowly lower the mass onto the spring until the spring is compressed to the point where it supports the mass at rest by itself and you can let go. How much is the spring compressed at that point? How much work did you do lowering the mass to that point?

b) Once again, hold the mass barely touching the top of the spring. This time you just let go of the mass. What is the magnitude of the spring's compression when the mass is at the lowest point it reaches, |slowest|? How much work did you do in this case?

c)This time, start with the mass sitting in equilibrium on the spring and push the mass down very slowly until it is at rest with the spring compressed s0. The spring is not attached to the end of the spring so it shoots high off the end of the spring when you let go of it. What is the mass' speed when it is 2L0 above the table? How much work did you do while compressing the spring?
2. Relevant equations
$ΔE = W$
$Δp = F_netΔt$
$K = \frac{1}{2}mv^2$
$U_g = mgh$ (for heights close to Earth's surface)
$U_{spring} = \frac{1}{2}k_ss^2$
3. The attempt at a solution
a)
System: mass,spring,Earth
Surroundings: hand
$F_g = mg$
$F_{spring} = -k_s|s|$
Δp = 0 because it is in equilibrium and Δt ≠ 0, so Fnet = 0
$F_{net} = F_g + F_{spring}$
$0 = (-mg - k_ss) \hat y$
Note: Even though the spring's force is upwards and should be positive I believe the sign here should be negative as the s would be negative meaning two negatives give a positive which is the correct result.
$s = \frac{mg}{k_s}$

$ΔE = W_{hand}$
$ΔK + ΔU_g + ΔU_spring = W_{hand}$
ΔK = 0 because it is at rest in the beginning and at the end and the initial potential energy of the spring = 0 as it was not stretched nor compressed.
$0 + (mg(L_0 - |s|) - mg(L_0)) + \frac{1}{2}k_ss^2-0 = W_{hand}$
$mgL_0 - mgs - mgL_0 + \frac{1}{2}k_ss^2 = W_{hand}$
$-1\frac{m^2g^2}{k_s} + \frac{1}{2}\frac{m^2g^2}{k_s} = W_{hand}$

$W_{hand} = \frac{-1}{2}\frac{m^2g^2}{k_s}$
I believe this is the correct answer for the first part.

b)
System: spring,mass,Earth
Surroundings: nothing
$ΔE = W$
$ΔK + ΔU_g + ΔU_{spring} = 0$
ΔK = 0 for the same reason as before, initial and final kinetic energies are 0.
$mg(L_0 - s_{lowest}) - mgL_0 + \frac{1}{2}k_ss_{lowest}^2 - 0 = 0$
$-mgs_{lowest} + \frac{1}{2}k_ss_{lowest}^2 = 0$
$s_{lowest}(-mg + \frac{1}{2}k_ss_{lowest}) = 0$

$s_{lowest} = \frac{2mg}{k_s}$
Since there are no external forces from the surroundings, especially that of your hand, the work done by your hand is 0.
$W_{hand} = 0$

c)
System: mass,spring,Earth
Surroundings: hand
$\frac{1}{2}k_ss_0^2 = U_{spring}$
$K + U = Constant$
Since at s0 the mass is not moving it has zero kinetic energy meaning that the constant is equal to the potential energy at s0.
$\frac{1}{2}mv_i^2 = \frac{1}{2}k_ss_0^2$
$v_i = \sqrt{\frac{k_ss_0^2}{m}} = s_0 \sqrt{frac{k_s}{m}}$
$W = ΔK$
$F_{net}⋅Δr = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
Fnet = -mg since air resistance is negligible
$-mg(2L_0-L_0) = \frac{1}{2}mv_f^2 - \frac{1}{2}m(s_0\sqrt{\frac{k_s}{m}})^2$
$-mgL_0 = \frac{1}{2}mv_f^2 - \frac{1}{2}s_0^2k_s$
$-mgL_0 + \frac{1}{2}s_0^2k_s = \frac{1}{2}mv_f^2$
$-2gL_0 + \frac{k_ss_0^2}{m} = v_f^2$

$v_f = \sqrt{\frac{k_ss_0^2}{m} - 2gL_0}$

$ΔE = W$
$ΔK + ΔU_g + ΔU_{spring} = W_{hand}$
$0 + mg(L_0 - s_0) - mg(L_0-s) + \frac{1}{2}k_ss_0^2 - \frac{1}{2}k_ss^2 = W_{hand}$
$-mgs_0 + mgs + \frac{1}{2}k_ss_0^2 - \frac{1}{2}k_ss^2 = W_{hand}$
$mg(s-s_0) + \frac{1}{2}k_s(s_0^2-s^2) = W_{hand}$
from previous problem....
$s = \frac{mg}{k_s}$
$mg(\frac{mg}{k_s} - s_0) + \frac{1}{2}k_s(s_0^2 - \frac{m^2g^2}{k_s^2}) = W_{hand}$
$\frac{m^2g^2}{k_s} - mgs_0 + \frac{1}{2}k_ss_0^2 - \frac{1}{2}\frac{m^2g^2}{k_s} = W_{hand}$

$\frac{1}{2}\frac{m^2g^2}{k_s} - s_0(mg + \frac{1}{2}k_ss_0) = W_{hand}$

2. Oct 22, 2014

### BvU

To be clear: PF is explicitly not in the business of stamp-approving homework. It would cause trouble with the teachers of this world, and they have plenty problems without us adding to them :)

I checked a and b, but I would violate the rules if I said I can't find anything wrong. In a), I am surprised at the "which is the correct result" . How do you know and doesn't that contradict the earlier "I have no way of checking" ?

In general, your way of checking if an answer is right or not is very simple: check if it satisfies the relevant equations. Sounds dumb but is really useful. Checks the math (and a significant part of the physics if you check the dimensions too.. .) It also shifts the burden to finding the right set of relevant equations to apply: that's Physics with a capital P.

I do have a few tips:

First, clearly define a coordinate system, and express your equations in that coordinate system. Properly include the signs (directions). This can avoid a lot of errors, and mostly it eliminates absolute values too. Forces, velocities, accelerations are vectors. Generally, y is upwards; this means acceleration from gravity $\vec g$ is -9.81 m/s2 in the positive y direction and change in potential energy is $mg(y-y_0)\$, in which g is a positive value...
In your work for a), I see $F_g = mg$ on one line, and $-mg$ four lines down...
If, in part a) s is negative, then in part c) you can't write $\sqrt{ks^2\over m} = s \sqrt{k\over m}$.
And "compression" means $\vec{ \Delta y}$ is pointing in the negative y direction. That way $\vec F_{spring} = -k_s \,\vec{\Delta y}\$ is pointing upwards.

Then: make sure it's clear (to yourself foremost) what the symbols stand for. For example, in your c) part, I have difficulty understanding the $_i$ in $v_i$. Can you explain ? And don't let anything fall out of the blue ($r$ ?).