Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 6.7.29a

Calculating work for different springs Calculate the work required to stretch the following springs 1.25 m from their equilibrium positions. Assume Hooke’s law is obeyed.
a. A spring that requires 100 J of work to be stretched 0.5 m from its equilibrium position

검증된 단계별 안내
1
Recall that the work done to stretch or compress a spring from its equilibrium position is given by the formula \(W = \frac{1}{2} k x^2\), where \(k\) is the spring constant and \(x\) is the displacement from equilibrium.
Use the information given for the first spring: it requires 100 J of work to stretch it 0.5 m. Substitute \(W = 100\) J and \(x = 0.5\) m into the formula to find the spring constant \(k\) by solving \(100 = \frac{1}{2} k (0.5)^2\).
Rearrange the equation to solve for \(k\): multiply both sides by 2 and divide by \((0.5)^2\) to isolate \(k\).
Once you have the value of \(k\), use it to calculate the work required to stretch the spring 1.25 m by substituting \(x = 1.25\) m into the work formula \(W = \frac{1}{2} k x^2\).
Evaluate the expression to find the work done for the 1.25 m stretch, which will give you the required answer.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
4m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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 linear relationship is fundamental for calculating forces and work in spring problems.
추천 영상:
가이드 코스
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 is calculated as the integral of force over distance: W = ∫ F dx. For springs, this results in W = (1/2) k x², representing the energy stored in the spring when stretched or compressed.
추천 영상:
가이드 코스
05:40
Work Done On A Spring (Hooke's Law)

Determining the Spring Constant from Work

Given the work done to stretch a spring a certain distance, the spring constant k can be found by rearranging the work formula: k = 2W / x². Knowing k allows calculation of work for other displacements, enabling solutions to problems involving different stretch lengths.
추천 영상:
가이드 코스
05:40
Work Done On A Spring (Hooke's Law)
관련 실천
교과서 질문

Displacement and distance from velocity Consider the graph shown in the figure, which gives the velocity of an object moving along a line. Assume time is measured in hours and distance is measured in miles. The areas of three regions bounded by the velocity curve and the t-axis are also given.

a. On what intervals is the object moving in the positive direction?

49
views
교과서 질문

Determine whether the following statements are true and give an explanation or counterexample. 


a. If the curve y=f(x) on the interval [a, b] is revolved about the y-axis, the area of the surface generated is ∫f(b)f(a) 2πf(y)√1+f′(y)^2 dy.

64
views
교과서 질문

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. When using the shell method, the axis of the cylindrical shells is parallel to the axis of revolution.

53
views
교과서 질문

Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi/hr). Assume t is measured in hours.

Theo: vT(t)=10, for t≥0

Sasha: vS(t)=15t, for 0≤t≤1, and vS(t)=15, for t>1


a. Graph the velocity function for both riders. 

28
views
교과서 질문

21–30. {Use of Tech} Arc length by calculator


a. Write and simplify the integral that gives the arc length of the following curves on the given interval. 

y = ln x, for 1≤x≤4

39
views
교과서 질문

For the given regions R₁ and R₂, complete the following steps.


a. Find the area of region R₁.


R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

34
views