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## Homework Statement

_{0}. the window is at height H.

1) What's the elongation of each of the spring pairs?

2) What's the total ladder's length.

3) The princess's lover, mass M, climbs to the first step and waits to the oscillations to stop. what is the ladder's length now.

4) The lover now climbs to the second step and waits. what's the length now.

5) When he passes from step 1 to 2, what's the change in the system's mechanic energy?

6) When he sits on the n-th step, what's the ladder's length?

7) If he would climb up slowly to the window, what would be the length of his way?

## Homework Equations

Equivalent spring of two springs in a row: ##\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}##

Equivalent spring of two parallel springs: ##k_{eq}=k_1+k_2##

$$\sum_1^n n=\frac{n(n+1)}{2}$$

## The Attempt at a Solution

1) The elongation of each of the n-th pair of strings: ##nmg=\overline k\cdot \Delta x_n~\rightarrow~\Delta x_n=\frac{nmg}{2k}##

2) The total length of n stages of the ladder, starting from the bottom:

$$L_n=n(l_0+d)+\sum \Delta x_n=n(l_0+d)+\frac{mg}{2k}\sum n=n\left[ l_0+d+\frac{mg(n+1)}{4k}\right]$$

Length of 20 steps:

$$L_{20}=20\left[ l_0+d+\frac{21mg}{4k} \right]$$

The distance of the n-th step from the window (considering only the step's mass m), ##L_{(20-n)m}##, is the reduction of the n stage's length, measured from the bottom, from the total length:

$$L_{(20-n)m}=\left( 20(l_0+d)+\sum_1^{20} \Delta x_n \right)-\left(n(l_0+d)+\sum_1^n \Delta x_n \right)=(20-n)(l_0+d)+205-\frac{n(n+1)}{2}$$

3) The equivalent spring constant of n pairs of parallel springs in a row is ##k_{eq}=\frac{2k}{n}##

The equivalent spring constant of the springs above the n-th step is ##k_{eq}=\frac{2k}{21-n}##

When M is on the first step:

$$L_{20m+M}=20\left[ l_0+d+\frac{21mg}{4k} \right]+\frac{10Mg}{k}$$

When M is on the n-th step it's distance from the window, ##L_{(20-n)M}##, is:

$$L_{(20-n)M}=L_{(20-N)m}+L_{nM}=(20-n) (l_0+d)+205-\frac{n(n+1)}{2}+\frac{Mg(21-n)}{2k}$$

$$h_{12}=L_{20+M}-L_{19+M}=20\left[ l_0+d+\frac{21mg}{4k} \right]+\frac{20Mg}{2k}-19\left[ l_0+d+\frac{5mg}{k} \right]-\frac{19Mg}{2k}=l_0+d+\left( 10m+\frac{M}{2} \right)\frac{g}{k}$$

$$\Delta h=\frac{Mg}{2k}$$

The energy gain:

$$\Delta E=Mg\cdot h_{12}$$

6) It doesn't matter on which step M sits, the expansion of the springs above it is the same as if he were sitting on the first one since the force is transmitted to all the steps. i mean, if he sits on the first step and i measure the elongation in the n-th set of springs, the same elongation would be if he sat on that n-th step.

So, the distance he crosses on his way up is just the total length of the ladder with him on the first step: