Answer to Question #145235 in Mechanics | Relativity for Aobakwe

Question

A linear spring has a free length of 56.6 cm. When it is under a hanging load of 22949N, its total length is 1214.3 cm.  

1. the spring constant k is  
2. the hanging load that makes it 1749.7 cm long is  
3. the energy stored in the spring when the extension is 1749.7 cm is  

Answer 1

According to the Hook's law,

$F = k \Delta l$

where F is load, k - stiffness and $\Delta l = l_0 - l$ - length difference.

1. $\displaystyle k = \frac{F}{\Delta l} = \frac{22949}{(1214.3 - 56.6) \cdot 10^{-2} }= 1982.3 \; \;\frac{N}{m}$
2. $F = 19 82.3 \cdot (1749.7-56.6) \cdot 10^{-2}= 33 562 \; N$
3. $\displaystyle E = \frac{k (\Delta l)^2}{2} = \frac{1982.3 \cdot (1749.7-56.6)^2 \cdot 10^{-4}}{2} = 284\,122\; J$