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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.7.60a
Chapter 6, Problem 6.7.60a

A vertical spring A 10-kg mass is attached to a spring that hangs vertically and is stretched 2 m from the equilibrium position of the spring. Assume a linear spring with F(x) = kx.
a. How much work is required to compress the spring and lift the mass 0.5 m?

Verified step by step guidance
1
Identify the forces and displacements involved: The mass stretches the spring 2 m from its equilibrium position, so the spring constant \( k \) can be found using Hooke's Law \( F = kx \), where \( F \) is the weight of the mass and \( x \) is the displacement.
Calculate the spring constant \( k \) by setting the force equal to the weight of the mass: \( F = mg = kx \), where \( m = 10 \) kg, \( g = 9.8 \) m/s\^2, and \( x = 2 \) m. Rearrange to find \( k = \frac{mg}{x} \).
Determine the total displacement for the work calculation: compressing the spring and lifting the mass by 0.5 m means the spring is compressed from its stretched position by 0.5 m, so the new displacement from equilibrium is \( 2 - 0.5 = 1.5 \) m.
Set up the integral for the work done in compressing the spring from 2 m to 1.5 m using the formula for work done by a variable force: \( W = \int_{x_1}^{x_2} F(x) \, dx = \int_{1.5}^{2} kx \, dx \).
Evaluate the integral to find the work done: integrate \( kx \) with respect to \( x \) over the limits from 1.5 to 2, which gives \( W = \left[ \frac{kx^2}{2} \right]_{1.5}^{2} \). This expression represents the work required to compress the spring and lift the mass.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a linear spring is proportional to its displacement from equilibrium, expressed as F(x) = kx, where k is the spring constant and x is the displacement. This relationship is fundamental for calculating forces and work involving springs.
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Work Done On A Spring (Hooke's Law)

Work Done by a Variable Force

When a force varies with position, the work done is found by integrating the force over the displacement. For a spring, work is calculated as the integral of F(x) = kx from the initial to final position, representing the energy required to compress or stretch the spring.
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Work Done On A Spring (Hooke's Law)

Gravitational Potential Energy

Lifting a mass against gravity increases its gravitational potential energy, calculated as mgh, where m is mass, g is acceleration due to gravity, and h is height. This energy must be included when determining the total work done in lifting the mass.
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Determining Concavity Given a Function
Related Practice
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a. Write and simplify the integral that gives the arc length of the following curves on the given interval. 

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Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).

a. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full?

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Textbook Question

17–22. Position from velocity Consider an object moving along a line with the given velocity v and initial position.


a. Determine the position function, for t≥0, using the antiderivative method


v(t) = 6−2t on [0, 5]; s(0)=0

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Textbook Question

Region R is revolved about the line x=4 to form a solid of revolution.


a. What is the radius of a cross section of the solid at a point y in [1, 3]?