Question

Forty-five identical springs are placed side by side (in parallel) and connected to a large massive block. The stiffness of the 45-spring combination is 20,250N/m. What is the stiffness of one of the individual springs?

Short answer

The value of stiffness of individual springs is 450 N/m

Step-by-step solution

Identification of given data

The given data is listed below,

• The number of the identical spring is, 45
• The stiffness of the springs is,$k_s=20,250\mathrm{N}/\mathrm{m}$

Significance of the stiffness of spring

Stiffness is described as the property of an object which mainly resists deformation if placed under an external force.

The expression for the stiffness of individual spring is expressed as,

$k_j=\frac{k_s}{N}$ …(i)

Here, is the stiffness of the one of the individual springs, is the stiffness of one spring and is the number of springs.

Calculation for the stiffness of the spring

Substitute all the values in the above equation.

$$\begin{gathered} K_i=\frac{20,250\mathrm{N}/\mathrm{m}}{45} \\ =450\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the stiffness of individual springs is 450 N/m.