Question

Five identical springs, each with stiffness $\mathbf{390}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$, are attached in parallel (that is, side by side) to hold up a heavy weight. If these springs were replaced by an equivalent single spring, what should be the stiffness of this single spring?

Short answer

The value of stiffness of one of the individual springs is$1950\mathrm{N}/\mathrm{m}$

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The number of the identical spring is, 5 .
• The stiffness of the springs is, ${\mathrm{K}}_{\mathrm{i}}=390\mathrm{N}/\mathrm{m}$

Significance of the stiffness

In the case of a parallel combination of springs, the stiffness of an individual spring is expressed as follows,

${\mathbf{K}}_{\mathbf{s}}\mathbf{=}{\mathbf{NK}}_{\mathbf{i}}$ …(i)

Here, ${\mathrm{K}}_{\mathrm{i}}$is the stiffness of the one of the individual springs, ${\mathrm{K}}_{\mathrm{s}}$is the stiffness of the single spring and $\mathrm{N}$is the number of springs.

Calculation for the stiffness of the spring

Substitute all the values in equation (i).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{s}}=390\mathrm{N}/\mathrm{m}\times 5 \\ =1950\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the stiffness of spring is $1950\mathrm{N}/\mathrm{m}$.