Textbook Question
13–20. Mass of one-dimensional objects Find the mass of the following thin bars with the given density function.
ρ(x) = 5e^-2x,for 0≤x≤4
37
views
10. Physics Applications of Integrals
Work
2:37 minutesProblem 6.R.73b
Textbook Question
Spring work
b. It takes 50 N of force to stretch a spring 0.2 m from its equilibrium position. How much work is needed to stretch it an additional 0.5 m?
Verified step by step guidance1
Identify the spring constant \( k \) using Hooke's Law, which states that the force \( F \) required to stretch or compress a spring by a distance \( x \) is \( F = kx \). Given \( F = 50 \) N and \( x = 0.2 \) m, solve for \( k \) by rearranging the formula to \( k = \frac{F}{x} \).
Express the work done to stretch a spring from position \( x = a \) to \( x = b \) using the integral formula for work: \( W = \int_a^b F(x) \, dx \), where \( F(x) = kx \) is the force at position \( x \).
Set the limits of integration to represent the additional stretch: since the spring is already stretched 0.2 m, the work needed to stretch it an additional 0.5 m corresponds to stretching from \( x = 0.2 \) m to \( x = 0.7 \) m (because \( 0.2 + 0.5 = 0.7 \)).
Write the integral for the work done as \( W = \int_{0.2}^{0.7} kx \, dx \). Substitute the value of \( k \) found in step 1 into this integral.
Evaluate the integral \( \int_{0.2}^{0.7} kx \, dx = k \int_{0.2}^{0.7} x \, dx = k \left[ \frac{x^2}{2} \right]_{0.2}^{0.7} \) to find the work done in stretching the spring the additional 0.5 m.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2mKey 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 required to stretch or compress a spring is proportional to the displacement from its equilibrium position, expressed as F = kx, where k is the spring constant and x is the displacement. This law helps determine the spring constant from the given force and displacement.
Recommended video:
05:40
Work Done On A Spring (Hooke's Law)
Work Done by a Variable Force
When stretching a spring, the force varies with displacement, so work done is calculated by integrating the force over the distance stretched. The work done to stretch a spring from x = a to x = b is W = (1/2)k(b² - a²), reflecting the area under the force-displacement curve.
Recommended video:
05:40Work Done On A Spring (Hooke's Law)
Spring Constant Calculation
The spring constant k quantifies the stiffness of the spring and is found by rearranging Hooke's Law: k = F/x. Knowing k allows calculation of work done for any displacement, making it essential to first find k using the initial force and displacement before solving for additional work.
Recommended video:
05:40Work Done On A Spring (Hooke's Law)
Related Videos
Related Practice