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Chapter 6: Problem 23

Calculate the work required to stretch the following springs \(0.5 \mathrm{m}\) from their equilibrium positions. Assume Hooke's law is obeyed. a. \(\mathrm{A}\) spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.2 \mathrm{m}\) from its equilibrium position b. A spring that requires \(50 \mathrm{J}\) of work to be stretched \(0.2 \mathrm{m}\) from its equilibrium position

Short Answer

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**Question:** Calculate the work required for stretching two different springs by 0.5 meters from their equilibrium positions. The force required to stretch the first spring by 0.2 meters is 50 N, while the work required to stretch the second spring by 0.2 meters is 50 J. **Answer:** The work required to stretch the first spring by 0.5 meters is 31.25 Joules and the work required to stretch the second spring by 0.5 meters is 312.5 Joules.

Step by step solution

01

Part a: Find the spring constant for the first spring

We are given the force required to stretch the first spring by 0.2 meters: \(F = 50 \mathrm{N}\) \(x = 0.2 \mathrm{m}\) Using Hooke's law, we can find the spring constant: \(k = \frac{F}{x} = \frac{50}{0.2} = 250 \frac{\mathrm{N}}{\mathrm{m}}\)
02

Part a: Calculate the work required to stretch the first spring

Now, we can use the calculated spring constant \(k\) to find the work required to stretch the spring by 0.5 meters: \(W = \frac{1}{2}kx^2 = \frac{1}{2}(250)(0.5)^2 = 31.25 \mathrm{J}\) So, the work required to stretch the first spring by 0.5 meters is 31.25 Joules.
03

Part b: Find the spring constant for the second spring

We are given the work required to stretch the second spring by 0.2 meters: \(W = 50 \mathrm{J}\) \(x = 0.2 \mathrm{m}\) We can use the work formula to find the spring constant \(k\): \(W = \frac{1}{2}kx^2\) Rearrange and solve for \(k\): \(k = \frac{2W}{x^2} = \frac{2(50)}{(0.2)^2} = 2500 \frac{\mathrm{N}}{\mathrm{m}}\)
04

Part b: Calculate the work required to stretch the second spring

Now, we can use the calculated spring constant \(k\) to find the work required to stretch the spring by 0.5 meters: \(W = \frac{1}{2}kx^2 = \frac{1}{2}(2500)(0.5)^2 = 312.5 \mathrm{J}\) So, the work required to stretch the second spring by 0.5 meters is 312.5 Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
When working with springs, the spring constant, denoted as \( k \), is a crucial aspect. It tells us how stiff a spring is. Hooke's Law relates the force exerted by a spring to its extension or compression, expressed as \( F = kx \), where \( F \) is the force applied, \( x \) is the displacement from equilibrium, and \( k \) is the spring constant.
To find \( k \), divide the force applied by the displacement it causes.
  • For example, if a spring requires 50 N to stretch 0.2 m, the spring constant \( k \) is calculated as \( k = \frac{50}{0.2} = 250 \, \mathrm{N/m} \).
  • This means for every meter the spring is stretched, it needs 250 N of force.
Understanding the spring constant helps in predicting how much force a spring will exert at different lengths.
Work Calculation
Calculating the work done on a spring involves understanding how much energy is needed to stretch or compress it from its equilibrium position. Using the work formula within the framework of Hooke鈥檚 Law, the work \( W \) is given by \( W = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement.
This quadratic relationship means that as you increase the displacement, the work increases quadratically.
  • For instance, if a spring with a spring constant \( k = 250 \, \mathrm{N/m} \) is stretched by 0.5 m, the work done is \( W = \frac{1}{2} \times 250 \times (0.5)^2 = 31.25 \mathrm{J} \).
  • Work is measured in joules (J), and conveys the energy transferred.
Thus, doubling the displacement requires four times the work, illustrating the importance of precise calculations.
Equilibrium Position
The equilibrium position is where the spring naturally rests without any force applied. It's the starting point from which measurements of displacement are taken.
A spring behaves according to Hooke鈥檚 Law, which assumes the force needed to change its position is directly proportional to the distance moved from this equilibrium.
  • In practical terms, this means that if you stretch a spring by a distance \( x \) from this position, the spring will try to return to equilibrium with a force \( F = kx \).
  • Understanding this concept is crucial in applications involving springs, as it dictates the balance of forces.
Maintaining other factors in constant, any movement from equilibrium requires a proportional force to counteract the spring's natural resistance to change.

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Most popular questions from this chapter

\(A\) 2000-liter cistern is empty when water begins flowing into it (at \(t=0\) ) at a rate (in \(\mathrm{L} / \mathrm{min}\) ) given by \(Q^{\prime}(t)=3 \sqrt{t},\) where \(t\) is measured in minutes. a. How much water flows into the cistern in 1 hour? b. Find and graph the function that gives the amount of water in the tank at any time \(t \geq 0\) c. When will the tank be full?

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A right circular cylinder with height \(R\) and radius \(R\) has a volume of \(V_{C}=\pi R^{3}\) (height \(=\) radius). a. Find the volume of the cone that is inscribed in the cylinder with the same base as the cylinder and height \(R\). Express the volume in terms of \(V_{C}\). b. Find the volume of the hemisphere that is inscribed in the cylinder with the same base as the cylinder. Express the volume in terms of \(V_{C}\).

Suppose a cylindrical glass with a diameter of \(\frac{1}{12} \mathrm{m}\) and a height of \(\frac{1}{10} \mathrm{m}\) is filled to the brim with a 400-Cal milkshake. If you have a straw that is 1.1 m long (so the top of the straw is \(1 \mathrm{m}\) above the top of the glass), do you burn off all the calories in the milkshake in drinking it? Assume that the density of the milkshake is \(1 \mathrm{g} / \mathrm{cm}^{3}(1 \mathrm{Cal}=4184 \mathrm{J})\)
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