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Applications of Integration

7.5 Work

Focus on variable-force work, lifting against gravity, spring equilibrium and bounds.

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Ideas in this lesson

01Variable-force workWorked steps · five questions02Lifting against gravityWorked steps · five questions03Spring equilibrium and boundsWorked steps · five questions
Idea 1 of 3

Variable-force work

Work is the integral of force along displacement, not necessarily final force times total distance.

Worked example 1

A spring has constant k=10Nmk = 10 \dfrac{N}{m}. Find the work in joules to stretch it from 00 to 0.2m0.2 m beyond equilibrium.
  1. F(x)=10xF(x) = 10x
  2. W=∫015≈0.210x dx=[10x22]∣015≈0.2W = \int_{0}^{\dfrac{1}{5} \approx 0.2} 10x\,dx = [\dfrac{10x^{2}}{2}]\big|_{0}^{\dfrac{1}{5} \approx 0.2}
  3. W=102(15≈0.22−02)=15≈0.2W = \dfrac{10}{2}(\dfrac{1}{5} \approx 0.2^{2} - 0^{2}) = \dfrac{1}{5} \approx 0.2

Answer: 15≈0.2\dfrac{1}{5} \approx 0.2

Worked example 2

A spring has constant k=20Nmk = 20 \dfrac{N}{m}. Find the work in joules to stretch it from .11 to 0.3m0.3 m beyond equilibrium.
  1. F(x)=20xF(x) = 20x
  2. W=∫0.10.320x dx=[20x22]∣0.10.3W = \int_{0.1}^{0.3} 20x\,dx = [\dfrac{20x^{2}}{2}]\big|_{0.1}^{0.3}
  3. W=202(0.32−0.12)=45≈0.8W = \dfrac{20}{2}(0.3^{2} - 0.1^{2}) = \dfrac{4}{5} \approx 0.8

Answer: 45≈0.8\dfrac{4}{5} \approx 0.8

Idea 2 of 3

Lifting against gravity

With constant weight, work is weight times height. Use consistent force and distance units.

Worked example 1

A 22 kg object is raised 1m1 m at constant speed. Use g=9.8g = 9.8 m/s2^{2}. Find the work in joules.
  1. F=F = mg =2⋅9.8=985≈19.6N= 2 \cdot 9.8 = \dfrac{98}{5} \approx 19.6 N
  2. W=W = Fh =985≈19.6⋅1J=985≈19.6= \dfrac{98}{5} \approx 19.6 \cdot 1 J = \dfrac{98}{5} \approx 19.6

Answer: 985≈19.6\dfrac{98}{5} \approx 19.6

Worked example 2

A 44 kg object is raised 2m2 m at constant speed. Use g=9.8g = 9.8 m/s2^{2}. Find the work in joules.
  1. F=F = mg =4⋅9.8=39.2N= 4 \cdot 9.8 = 39.2 N
  2. W=W = Fh =39.2⋅2J=3925≈78.4= 39.2 \cdot 2 J = \dfrac{392}{5} \approx 78.4

Answer: 3925≈78.4\dfrac{392}{5} \approx 78.4

Idea 3 of 3

Spring equilibrium and bounds

Stretch is measured from equilibrium. Changing the starting stretch changes the lower integral bound.

Worked example 1

A spring has constant k=10Nmk = 10 \dfrac{N}{m}. Find the work in joules to stretch it from 00 to 0.2m0.2 m beyond equilibrium.
  1. F(x)=10xF(x) = 10x
  2. W=∫015≈0.210x dx=[10x22]∣015≈0.2W = \int_{0}^{\dfrac{1}{5} \approx 0.2} 10x\,dx = [\dfrac{10x^{2}}{2}]\big|_{0}^{\dfrac{1}{5} \approx 0.2}
  3. W=102(15≈0.22−02)=15≈0.2W = \dfrac{10}{2}(\dfrac{1}{5} \approx 0.2^{2} - 0^{2}) = \dfrac{1}{5} \approx 0.2

Answer: 15≈0.2\dfrac{1}{5} \approx 0.2

Worked example 2

A spring has constant k=20Nmk = 20 \dfrac{N}{m}. Find the work in joules to stretch it from .11 to 0.3m0.3 m beyond equilibrium.
  1. F(x)=20xF(x) = 20x
  2. W=∫0.10.320x dx=[20x22]∣0.10.3W = \int_{0.1}^{0.3} 20x\,dx = [\dfrac{20x^{2}}{2}]\big|_{0.1}^{0.3}
  3. W=202(0.32−0.12)=45≈0.8W = \dfrac{20}{2}(0.3^{2} - 0.1^{2}) = \dfrac{4}{5} \approx 0.8

Answer: 45≈0.8\dfrac{4}{5} \approx 0.8

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.5,p7.5, p. 477477. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.