# Suppose that 6 J of work is needed to stretch a spring from its natural length of 36 cm to a...

## Question:

Suppose that 6 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 51 cm.

A. How much work is needed to stretch the spring from 44 cm to 49 cm?

B. How far beyond its natural length will a force of 40 N keep the spring stretched?

## Hooke's Law

When a force is applied on an object, it can cause a change in the shape of the object. The amount of change in shape or deformation {eq}\Delta x {/eq} induced by the applied force {eq}F {/eq} is given by the Hooke's law:

$$F=-k\Delta x $$,

where {eq}k {/eq} is a proportionality constant that depends upon the property of the object and the applied force.

## Answer and Explanation:

**We are given:**

- Natural length of the spring, {eq}L_n=36\ \rm {cm} {/eq}

- Change in the length of the spring when stretched, {eq}\Delta L=(51-36)\ \rm{cm}=15\ \rm{cm}=0.15\ \rm m {/eq}

- Work needed to stretch the spring by the given {eq}\Delta L {/eq}, {eq}W=6\ \rm J {/eq}

**Part A** We are asked to calculate the work needed to stretch the spring from 44 cm to 49 cm.

**Solution:**

**Step I** To solve this problem, first we need to determine the spring constant {eq}k
{/eq} of the spring.

The work {eq}W {/eq} needed to be done to stretch the spring (spring constant, {eq}k {/eq}) by length {eq}\Delta L {/eq} is given by:

$$\begin{align} W=\frac{1}{2}k\Delta L^2 \end{align} $$

Using the given values of {eq}\Delta L {/eq} and {eq}W {/eq} we will calculate the spring constant:

$$\begin{align} k&=\frac{2W}{\Delta L^2}\\ &=\frac{2 \times 6\ \rm J}{(0.15\ \rm{m})^2}\\ &=533 \ \rm{N/m}\\ \end{align} $$

**Step II** Using the calculated value of the spring constant {eq}k
{/eq}, we can calculate the work needed to stretch the spring from 44 cm to 49 cm.

Change in the length of the spring when it is stretched from 44 cm to 49 cm, {eq}\Delta L_1=(49-44)\ \rm{cm}=5\ \rm{cm}=5 \times 10^{-2}\ \rm m {/eq}

Using the equation for the work done on the spring, we get:

$$\begin{align} W_1&=\frac{1}{2}k\Delta L_1^2\\ &=\frac{1}{2} \times 533 \ \rm{N/m} \times (5 \times 10^{-2}\ \rm m)^2\\ &=0.67 \ \rm J \\ \end{align} $$

**Hence, the work needed to stretch the spring from 44 cm to 49 cm is {eq}0.67\ \rm{J}
{/eq}.**

**Part B** We are asked to determine the change in the natural length of the spring induced by a force of 40 N.

**Solution:** The change in the natural length of the spring induced by a force of 40 N can be calculated using Hooke's law:

$$F_2=k\Delta L_2 $$

Using {eq}F_2=40\ \rm N {/eq} and the spring constant calculated in Part A, we get:

$$\begin{align} \Delta L_2&=\frac{F_2}{k}\\ &=\frac{40\ \rm N}{533 \ \rm{N/m}}\\ &=7.5 \times 10^{-2}\ \rm m \\ &=7.5 \ \rm{cm} \end{align} $$

Length of the stretched spring, {eq}L_2=\Delta L_2 + L_n = (7.5+36) \ \rm{cm}=\mathbf{43.5 \ \rm{cm}} {/eq}

**Hence, when a force of 40 N is applied, it will stretch the spring by 7.5 cm, i.e., from its natural length of 36 cm to 43.5 cm.**

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.