- #1
The Wanderer
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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. Homework Statement
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?
[itex]ΔE = W[/itex]
[itex]Δp = F_netΔt[/itex]
[itex]K = \frac{1}{2}mv^2[/itex]
[itex]U_g = mgh[/itex] (for heights close to Earth's surface)
[itex]U_{spring} = \frac{1}{2}k_ss^2[/itex]
a)
System: mass,spring,Earth
Surroundings: hand
[itex]F_g = mg[/itex]
[itex]F_{spring} = -k_s|s|[/itex]
Δp = 0 because it is in equilibrium and Δt ≠ 0, so Fnet = 0
[itex]F_{net} = F_g + F_{spring}[/itex]
[itex]0 = (-mg - k_ss) \hat y[/itex]
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.
[itex]s = \frac{mg}{k_s}[/itex]
[itex]ΔE = W_{hand}[/itex]
[itex]ΔK + ΔU_g + ΔU_spring = W_{hand}[/itex]
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.
[itex]0 + (mg(L_0 - |s|) - mg(L_0)) + \frac{1}{2}k_ss^2-0 = W_{hand}[/itex]
[itex]mgL_0 - mgs - mgL_0 + \frac{1}{2}k_ss^2 = W_{hand}[/itex]
[itex]-1\frac{m^2g^2}{k_s} + \frac{1}{2}\frac{m^2g^2}{k_s} = W_{hand}[/itex]
[itex]W_{hand} = \frac{-1}{2}\frac{m^2g^2}{k_s}[/itex]
I believe this is the correct answer for the first part.
b)
System: spring,mass,Earth
Surroundings: nothing
[itex]ΔE = W[/itex]
[itex]ΔK + ΔU_g + ΔU_{spring} = 0[/itex]
ΔK = 0 for the same reason as before, initial and final kinetic energies are 0.
[itex]mg(L_0 - s_{lowest}) - mgL_0 + \frac{1}{2}k_ss_{lowest}^2 - 0 = 0[/itex]
[itex]-mgs_{lowest} + \frac{1}{2}k_ss_{lowest}^2 = 0[/itex]
[itex]s_{lowest}(-mg + \frac{1}{2}k_ss_{lowest}) = 0[/itex]
[itex]s_{lowest} = \frac{2mg}{k_s}[/itex]
Since there are no external forces from the surroundings, especially that of your hand, the work done by your hand is 0.
[itex]W_{hand} = 0[/itex]
c)
System: mass,spring,Earth
Surroundings: hand
[itex]\frac{1}{2}k_ss_0^2 = U_{spring}[/itex]
[itex]K + U = Constant[/itex]
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.
[itex]\frac{1}{2}mv_i^2 = \frac{1}{2}k_ss_0^2[/itex]
[itex]v_i = \sqrt{\frac{k_ss_0^2}{m}} = s_0 \sqrt{frac{k_s}{m}}[/itex]
[itex]W = ΔK[/itex]
[itex]F_{net}⋅Δr = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/itex]
Fnet = -mg since air resistance is negligible
[itex]-mg(2L_0-L_0) = \frac{1}{2}mv_f^2 - \frac{1}{2}m(s_0\sqrt{\frac{k_s}{m}})^2[/itex]
[itex]-mgL_0 = \frac{1}{2}mv_f^2 - \frac{1}{2}s_0^2k_s[/itex]
[itex]-mgL_0 + \frac{1}{2}s_0^2k_s = \frac{1}{2}mv_f^2[/itex]
[itex]-2gL_0 + \frac{k_ss_0^2}{m} = v_f^2[/itex]
[itex]v_f = \sqrt{\frac{k_ss_0^2}{m} - 2gL_0}[/itex]
[itex]ΔE = W[/itex]
[itex]ΔK + ΔU_g + ΔU_{spring} = W_{hand}[/itex]
[itex]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}[/itex]
[itex]-mgs_0 + mgs + \frac{1}{2}k_ss_0^2 - \frac{1}{2}k_ss^2 = W_{hand}[/itex]
[itex]mg(s-s_0) + \frac{1}{2}k_s(s_0^2-s^2) = W_{hand}[/itex]
from previous problem...
[itex]s = \frac{mg}{k_s}[/itex]
[itex]mg(\frac{mg}{k_s} - s_0) + \frac{1}{2}k_s(s_0^2 - \frac{m^2g^2}{k_s^2}) = W_{hand}[/itex]
[itex]\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}[/itex]
[itex]\frac{1}{2}\frac{m^2g^2}{k_s} - s_0(mg + \frac{1}{2}k_ss_0) = W_{hand}[/itex]
1. Homework Statement
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?
Homework Equations
[itex]ΔE = W[/itex]
[itex]Δp = F_netΔt[/itex]
[itex]K = \frac{1}{2}mv^2[/itex]
[itex]U_g = mgh[/itex] (for heights close to Earth's surface)
[itex]U_{spring} = \frac{1}{2}k_ss^2[/itex]
The Attempt at a Solution
a)
System: mass,spring,Earth
Surroundings: hand
[itex]F_g = mg[/itex]
[itex]F_{spring} = -k_s|s|[/itex]
Δp = 0 because it is in equilibrium and Δt ≠ 0, so Fnet = 0
[itex]F_{net} = F_g + F_{spring}[/itex]
[itex]0 = (-mg - k_ss) \hat y[/itex]
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.
[itex]s = \frac{mg}{k_s}[/itex]
[itex]ΔE = W_{hand}[/itex]
[itex]ΔK + ΔU_g + ΔU_spring = W_{hand}[/itex]
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.
[itex]0 + (mg(L_0 - |s|) - mg(L_0)) + \frac{1}{2}k_ss^2-0 = W_{hand}[/itex]
[itex]mgL_0 - mgs - mgL_0 + \frac{1}{2}k_ss^2 = W_{hand}[/itex]
[itex]-1\frac{m^2g^2}{k_s} + \frac{1}{2}\frac{m^2g^2}{k_s} = W_{hand}[/itex]
[itex]W_{hand} = \frac{-1}{2}\frac{m^2g^2}{k_s}[/itex]
I believe this is the correct answer for the first part.
b)
System: spring,mass,Earth
Surroundings: nothing
[itex]ΔE = W[/itex]
[itex]ΔK + ΔU_g + ΔU_{spring} = 0[/itex]
ΔK = 0 for the same reason as before, initial and final kinetic energies are 0.
[itex]mg(L_0 - s_{lowest}) - mgL_0 + \frac{1}{2}k_ss_{lowest}^2 - 0 = 0[/itex]
[itex]-mgs_{lowest} + \frac{1}{2}k_ss_{lowest}^2 = 0[/itex]
[itex]s_{lowest}(-mg + \frac{1}{2}k_ss_{lowest}) = 0[/itex]
[itex]s_{lowest} = \frac{2mg}{k_s}[/itex]
Since there are no external forces from the surroundings, especially that of your hand, the work done by your hand is 0.
[itex]W_{hand} = 0[/itex]
c)
System: mass,spring,Earth
Surroundings: hand
[itex]\frac{1}{2}k_ss_0^2 = U_{spring}[/itex]
[itex]K + U = Constant[/itex]
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.
[itex]\frac{1}{2}mv_i^2 = \frac{1}{2}k_ss_0^2[/itex]
[itex]v_i = \sqrt{\frac{k_ss_0^2}{m}} = s_0 \sqrt{frac{k_s}{m}}[/itex]
[itex]W = ΔK[/itex]
[itex]F_{net}⋅Δr = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/itex]
Fnet = -mg since air resistance is negligible
[itex]-mg(2L_0-L_0) = \frac{1}{2}mv_f^2 - \frac{1}{2}m(s_0\sqrt{\frac{k_s}{m}})^2[/itex]
[itex]-mgL_0 = \frac{1}{2}mv_f^2 - \frac{1}{2}s_0^2k_s[/itex]
[itex]-mgL_0 + \frac{1}{2}s_0^2k_s = \frac{1}{2}mv_f^2[/itex]
[itex]-2gL_0 + \frac{k_ss_0^2}{m} = v_f^2[/itex]
[itex]v_f = \sqrt{\frac{k_ss_0^2}{m} - 2gL_0}[/itex]
[itex]ΔE = W[/itex]
[itex]ΔK + ΔU_g + ΔU_{spring} = W_{hand}[/itex]
[itex]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}[/itex]
[itex]-mgs_0 + mgs + \frac{1}{2}k_ss_0^2 - \frac{1}{2}k_ss^2 = W_{hand}[/itex]
[itex]mg(s-s_0) + \frac{1}{2}k_s(s_0^2-s^2) = W_{hand}[/itex]
from previous problem...
[itex]s = \frac{mg}{k_s}[/itex]
[itex]mg(\frac{mg}{k_s} - s_0) + \frac{1}{2}k_s(s_0^2 - \frac{m^2g^2}{k_s^2}) = W_{hand}[/itex]
[itex]\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}[/itex]
[itex]\frac{1}{2}\frac{m^2g^2}{k_s} - s_0(mg + \frac{1}{2}k_ss_0) = W_{hand}[/itex]