Chapter 18, Problem 26A

A certain spring stretches 1 cm for each kilogram it supports.

a. If the elastic limit is not reached, how far will it stretch when it supports a load of 8 kg?

b. Suppose the spring is placed next to an identical spring so the two side-by-side springs equally share the 8-kg load. How far will each spring stretch?

To determine

The stretched distance of the spring when it supports a given load.

Answer to Problem 26A

The stretched distance of the spring is $8\text{ cm}$ .

Explanation of Solution

Given:

For each kilogram, spring stretches 1 cm.

Load is 8 kg.

Formula used:

Hooke’s law states that the amount of stretch and applied force are directly proportional to each other. The expression for Hooke’s law is

  $F\sim \Delta x$

Here, $\Delta x$ is the amount of stretch.

Elastic limit is the distance at which permanent distortion occurs.

Calculation:

A spring stretches $1\text{ cm}$ for each kilogram of supports.

The spring is supported by a load of $8\text{ kg}$ .

According to Hooke’s law, the spring will stretch at a distance of $8\text{ cm}$ when it supports a load of $8\text{ kg}$ .

Conclusion:

Thus, the stretched distance of the spring is $8\text{ cm}$ .

To determine

The stretched distance of the two side-by-side springs.

Answer to Problem 26A

The stretched distance of the two side-by-side spring is $4\text{ cm}$ .

Explanation of Solution

Given:

For each kilogram, spring stretches 1 cm.

Load is 8 kg.

A load is stretched by two side-by-side springs.

Formula used:

Hooke’s law states that the amount of stretch and applied force are directly proportional to each other. The expression for Hooke’s law is,

  $F\sim \Delta x$

Here, $\Delta x$ is the amount of stretch.

Elastic limit is the distance at which permanent distortion occurs.

A spring stretches $1\text{ cm}$ for each kilogram of supports.

The two side-by-side springs equally shared by a load of $8\text{ kg}$ .

Each of the spring is supported by a load of $4\text{ kg}$ .

According to Hooke’s law, each spring will stretch at a distance of $4\text{ cm}$ when it supports a load of $4\text{ kg}$ .

Conclusion:

Thus, the stretched distance of the each side-by-side spring is $4\text{ cm}$ .