Chapter 18, Problem 12A

A 2-kg mass stretches a spring 3 cm. How far does the spring stretch when it supports 6 kg? (Assume the spring has not reached its elastic limit.)

To determine

The stretch in spring.

Answer to Problem 12A

The extension produced by 6 kg weight is $9\text{cm}$ .

Explanation of Solution

Given:

Mass attached to spring is 2kg

Stretch in spring is 3cm

New mass attached is 6 kg.

Formula used:

Hook's law states that when a deforming force is applied on an object within elastic limits then extension/compression produced in the object is directly proportional to deforming force applied on the object.

  $F\alpha x$

  $F=kx$

Here, $F$ represents force applied on the object, $x$ represents extension/compression produced in the object, and $k$ is constant of proportionality.

Calculation:

A two kilogram mass attached to the spring and it stretches the spring $3\text{cm}$ .

Convert the extension from $\text{cm}$ into $\text{m}$ .

  $x=(3\text{cm})\left(\frac{1\text{m}}{100\text{cm}}\right)$

  $=0.03\text{m}$

The force acting on the spring is,

  $F=mg$

Rearrange the equation for Hook's law for $k$ .

  $k=\frac{F}{x}$

  $=\frac{mg}{x}$

Substitute $2\text{kg}$ for $m,9.8\text{m}/{\text{s}}^2$ for $g,$ and $3\text{cm}$ for $x$ .

  $k=\frac{(2\text{kg})\left(9.8\text{m}/{\text{s}}^2\right)}{0.03\text{m}}$

  $=653.33\text{N}/\text{m}$

When 6 kg mass attached to the spring, then stretch in the spring will be calculated by substituting $6\text{kg}$ for $m,9.8\text{m}/{\text{s}}^2$ for $g,$ and $653.33\text{N}/\text{m}$ for $k$ .

  $x=\frac{(6\text{kg})\left(9.8\text{m}/{\text{s}}^2\right)}{653.33\text{N}/\text{m}}$

  $=0.09\text{m}$

  $=(0.09\text{m})\left(\frac{100\text{cm}}{1\text{m}}\right)$

  $=9.0\text{cm}$

Conclusion:

Therefore, the extension produced by 6 kg weight is $9\text{cm}$ .