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## Homework Statement

A mass [tex]m[/tex] hangs on a vertical spring of spring constant [tex]k[/tex].

(a) How far will this hanging mass have stretched the spring from its relaxed length?

(b) If you now push up on the mass and lift it until the spring reaches its relaxed length, how much work will you have done against gravity? Against the spring?

## Homework Equations

I'll include them in my attempt at a solution.

## The Attempt at a Solution

(a)

[tex] F = -kx [/tex] <--- force of a spring

[tex] U(x) = -W [/tex]

[tex] U(x) = -\int_0^{x} -kx' dx'[/tex]

[tex] U(x) = k (\frac{1}{2}x'^2)_0^{x}[/tex]

[tex] U(x) = \frac{1}{2} kx^2[/tex] <--- potential energy of spring

The energy in the relaxed state is:

[tex]E_1 = \frac{1}{2}mv_1^{2} + \frac{1}{2}kx_1^{2} = \frac{1}{2}mv_1^{2} + 0 [/tex]

The energy in the stretched state is:

[tex]E_2 = \frac{1}{2}mv_2^{2} + \frac{1}{2}kx_2^{2} = 0 + \frac{1}{2}k(-x)^2[/tex]

The energy conserves, so [tex] E_1 = E_2 [/tex]:

[tex]\frac{1}{2}mv_1^{2} = \frac{1}{2}kx^2 [/tex]

[tex] x = \sqrt{\frac{mv_1^{2}}{k}}[/tex] <--- How far the spring stretches

(b)

[tex]U(0) - U(x) = -W [/tex]

[tex]-\frac{1}{2}kx^2 = -W [/tex]

[tex] W_{you} = \frac{1}{2}kx^2[/tex]

[tex] W_{grav} = -mg(-x) = mgx[/tex]

The work you do against gravity is [tex] W_{you} - W_{grav} [/tex]:

[tex] W = \frac{1}{2}kx^2 - mgx[/tex]

[tex] W = x ( \frac{1}{2}kx - mg) [/tex] <--- work you do against gravity

The work the spring does is:

[tex] W_{spring} = -\frac{1}{2}kx^2 [/tex]

[tex] W_{you} = \frac{1}{2}kx^2[/tex]

The work you do against the spring is [tex] W_{you} - W_{spring} [/tex] :

[tex]W = \frac{1}{2}kx^2 + \frac{1}{2}kx^2 [/tex]

[tex]W= kx^2 [/tex] <--- work you do against the spring

I have no idea if this is correct, as the answer is not in the back of the book.. so I can't check myself on this particular problem. Please critique my work. I don't know if all of the equations make sense, as I may have been just writing stuff down and trying to force things to work. Thanks in advance.