How much work is done by a person lifting a bucket off the ground?
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
10. Physics Applications of Integrals
Work
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
A spring requires a force of to stretch the spring to past its equilibrium point. How much work could it take to stretch the spring from to past equilibrium?
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Step 1: Recall Hooke's Law, which states that the force required to stretch or compress a spring is proportional to the displacement from its equilibrium position. Mathematically, this is expressed as F = kx, where F is the force, k is the spring constant, and x is the displacement.
Step 2: Use the given information to calculate the spring constant k. The problem states that a force of 8 N is required to stretch the spring 5 cm (0.05 m). Substitute these values into Hooke's Law: F = kx → 8 = k(0.05). Solve for k.
Step 3: The work done to stretch a spring is given by the formula W = ∫ F dx = ∫ kx dx, where the limits of integration are the initial and final displacements. In this case, the limits are 2 cm (0.02 m) and 9 cm (0.09 m). Substitute the value of k obtained in Step 2 into the integral.
Step 4: Evaluate the integral W = ∫ kx dx from 0.02 to 0.09. This involves finding the antiderivative of kx, which is (1/2)kx², and then applying the limits of integration.
Step 5: After evaluating the integral, the result will give the total work done to stretch the spring from 2 cm to 9 cm past its equilibrium point. Compare this value to the provided answer choices to identify the correct one.
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