Chapter 5: Applications of Integration
Section 5.8: Work
Example 5.8.2
A spring obeying Hooke's Law is extended to a total length of 5 in. The work done in extending it beyond this length to a total length of 10 in is 25 in-lbs. The work done in extending it further to a total length of 14 in is an additional 40 in-lbs. Find the natural length of the spring.
Solution
Mathematical Solution
Let L be the natural length of the spring (in inches), and let k be the spring constant (with units lbs/in). The work done stretching the spring from 5 to 10 inches is
∫5−L10−Lk x ⅆx=12k x2x=5−Lx=10−L=5 k152−L=25
The work done stretching the spring from 10 to 14 inches is
∫10−L14−Lk x ⅆx=4 k 12−L=40
Solving these two resulting equations gives L = 3, and k=10/9.
Maple Solution
Expression palette: Definite-integral template As per the discussion above, enter the equation that governs the work done in stretching the spring from 5 to 10 inches.
Press the Enter key.
∫5−L10−Lk x ⅆx=25
12k10−L2−5−L2=25
Expression palette: Definite-integral template As per the discussion above, enter the equation that governs the work done in stretching the spring from 10 to 14 inches.
∫10−L14−Lk x ⅆx=40
12k14−L2−10−L2=40
Using equation labels, construct a sequence of these two equations. Press the Enter key.
Context Panel: Solve≻Solve
,
12k10−L2−5−L2=25,12k14−L2−10−L2=40
→solve
L=3,k=109
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