Question

If a chain of 50identical short springs linked end to end has a stiffness of 270N/m, what is the stiffness of one short spring?

Short answer

The stiffness of the one short spring is $1.35\times {10}^4\mathrm{N}/\mathrm{m}$.

Step-by-step solution

Identification of given data

The given is listed as follows:

• Number of springs is 50
• The stiffness of the spring is,$k_t=270\mathrm{N}/\mathrm{m}$

Significance of the stiffness of spring

The stiffness of a spring is described as the ability of a particular material for resisting the deformation of a particular object.

The concept of the stiffness gives the stiffness of one short spring.

Calculation for the stiffness of one short spring

When the springs are calculated in the series form then the equation of the stiffness is expressed as follows:

$\frac{1}{k_t}=\frac{1}{k}+..............+\frac{50}{k}$

Here, $k_t$is the total stiffness of the spring and k is the stiffness of one spring.

Substitute all the values in the above equation.

$$\begin{gathered} \frac{1}{270\mathrm{N}/\mathrm{m}}=\frac{50}{k} \\ k=1.35\times {10}^4\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the stiffness on one spring is $1.35\times {10}^4\mathrm{N}/\mathrm{m}$.