Chapter 11, Problem 15P

(a)

To determine

To Find: The stiffness constant of spring.

Answer to Problem 15P

The stiffness constant of spring is $417\text{ N/m}$ .

Explanation of Solution

Given:

Work $W=3\text{ J}$ .

Amplitude $A=\text{0}.12\text{ m}$ .

Maximum acceleration $a_{\max }=15{\text{ m/s}}^{\text{2}}$ .

Formula used:

The stiffness constant of the spring is determined by the following formula,

  $W=\frac{1}{2}\times k\times x^2$

(Work done to compress the spring is called Potential energy of spring)

Here,

  $W$ is work done on spring,

  $x$ is the displacement of spring,

  $k$ is stiffness constant of spring.

Calculation:

From the above formula, the expression for spring constant k is derived by following way.

  $\begin{array}{l}W=\frac{1}{2}\times k\times x^2\\ k=\frac{2\times W}{x^2}=\frac{2\times 3}{{(0.12)}^2}=417\text{ N/m}\end{array}$

Conclusion:

Thus, the stiffness constant of spring is $417\text{ N/m}$ .

(b)

To determine

To Find: The mass of the block.

Answer to Problem 15P

The mass of the block is $3.3\text{ kg}$ .

Explanation of Solution

Given:

Spring constant $k=417\text{ N/m}$ .

Amplitude $A=\text{0}.12\text{ m}$ .

Maximum acceleration $a_{\max }=15{\text{ m/s}}^{\text{2}}$ .

Formula used:

Mass of the block given its direction and distance of the spring was compressed becomes the amplitude of the motion, so the mass of the block is determined by the following formula,

  $a_{\max }=\frac{k}{m}\times A$

Here,

  $m$ is mass on spring,

  $k$ is spring constant,

  $A$ is the amplitude of spring.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}m=\frac{k\times A}{a_{\max }}\\ m=\frac{417\times 0.12}{15}\\ \text{mass of block }m=3.3\text{ kg}\end{array}$

Conclusion:

Thus, the mass of the block is $3.3\text{ kg}$ .