Question

a uniform elastic string has length a1 when tension is t1 and length a2 when tension is t2 . the amount of work done in streching it from its natural length to length a1+a2

Answer 1

Answer on Question #62600, Physics / Mechanics | Relativity

A uniform elastic string has length $a1$ when tension is $t1$ and length $a2$ when tension is $t2$ . The amount of work done in stretching it from its natural length-to-length $a1 + a2$ .

Solution:

$\mathrm{T}_{1} = \mathrm{K}$ $(\mathrm{a}_{1} - \mathrm{a}_{0})$

$\mathrm{T}_{2} = \mathrm{K}$ (a2-a0)

K - Force constant of s string

$a_0$ - original length

$\mathrm{T}_{1} - \mathrm{T}_{2} = \mathrm{K}\mathrm{a}_{1} - \mathrm{K}\mathrm{a}_{2}$

$\mathrm{K} = \mathrm{T}_{1} - \mathrm{T}_{2} / \mathrm{a}_{1} - \mathrm{a}_{2}$

$\mathrm{T}_{1} = \left(\mathrm{T}_{1} - \mathrm{T}_{2} / \mathrm{a}_{1} - \mathrm{a}_{2}\right) \times \left(\mathrm{a}_{1} - \mathrm{a}_{0}\right)$

$a_0 = a_1 - \left(T_1(a_1 - a_2) / (T_1 - T_2)\right)$

$a_0 = T_1a_1 - T_2a_1 - T_1a_1 + T_1a_2 / (T_1 - T_2)$

$a_0 = T_1a_2 - T_2a_1 / (T_1 - T_2)$

$a_3 = a_1 + a_2$

$W = \int_{a_0}^{a_3}F(a)da = \int_{a_0}^{a_3}(ka)da = \frac{1}{2} ka_3^2 -\frac{1}{2} ka_0^2$

$W = \frac{1}{2} k(a_{3}^{2} - a_{0}^{2}) = \frac{1}{2}\frac{(a_{1}T_{1} - a_{2}T_{2})}{T_{1} - T_{2}(a_{1} - a_{2})}^{2}$

Answer: $\frac{1}{2}\frac{(a_1T_1 - a_2T_2)}{T_1 - T_2(a_1 - a_2)}$

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