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AP Calculus AB: 10.8.2 Calculating Work

Mathematics12 CardsCreated 3 months ago

This content explains how to calculate work done by forces, including constant and variable forces, using multiplication or definite integrals. It covers unit considerations and provides example problems involving forces acting over specified distances.

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Calculating Work

  • Work is the energy used when applying a force over a distance.

  • For a changing or variable force F(x) on [a, b], work is given by the integral

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Key Terms

Term
Definition

Calculating Work

  • Work is the energy used when applying a force over a distance.

  • For a changing or variable force F(x) on [a, b], work is give...

note

  • Work is the energy used to move an object. If the object doesn’t move in the direction of the force, then no work is done.

  • F...

A force described by F (x) = x − 2 lb (where x > 0) acts between x = 2 ft and x = k ft. A total of 162 lb-ft of work is done. What is k ?

20 ft

A refrigerator is pushed 10 feet across a kitchen with a constant force of 50 lb. How much work is done moving the refrigerator?

500 ft-lb

A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum. How much work is done by the crane?

18,000 N-m

Which results in the greatest amount of work?

Applying a force of F (x) = x N over a distance of 8 m

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AP Calculus AB: 10.8.2 Calculating Work
TermDefinition

Calculating Work

  • Work is the energy used when applying a force over a distance.

  • For a changing or variable force F(x) on [a, b], work is given by the integral

note

  • Work is the energy used to move an object. If the object doesn’t move in the direction of the force, then no work is done.

  • Finding the work done by a constant force is pretty simple. You just multiply the magnitude of the force by the change in distance.

  • Work is measured in force units times distance units. Be careful when talking about work. You have to make sure you are using the right units.

  • For a variable force, work is computed using a definite
    integral.

  • Consider this example. Remember that if Prof. Burger didn’t move the refrigerator, he would have done no work no matter how hard he pushed.

  • The force Prof. Burger is using to move the refrigerator is a variable force defined in terms of x.

  • To calculate the work, plug the force and the positions into the work formula.

  • Remember, the natural log of one is zero.

  • The force unit is the pound (lb). The distance unit is the foot (ft). Thus, the work unit is the foot-pound (ft-lb).

A force described by F (x) = x − 2 lb (where x > 0) acts between x = 2 ft and x = k ft. A total of 162 lb-ft of work is done. What is k ?

20 ft

A refrigerator is pushed 10 feet across a kitchen with a constant force of 50 lb. How much work is done moving the refrigerator?

500 ft-lb

A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum. How much work is done by the crane?

18,000 N-m

Which results in the greatest amount of work?

Applying a force of F (x) = x N over a distance of 8 m

What is the work done by a force described by F=100sinθlb over the distance betweenθ=0 and θ=π ft?

200 lb-ft

What is the work done by a force described by F = 8x^ 3 + 6x^ 2 + 4x + 2 N over the distance between x = 0 and x = 10 m?

22,220 N-m

Evaluate the following as true or false:
The work done by a varying force F over a distance D is equal to the work done by the average value of the force F held constant over the entire distance D.

true

Which results in the least amount of work?

Applying the force F = 4x lb over the distance from x = 0 to x = 3.

Which of the following is not an acceptable unit for work?

kg-cm

What is the work done by a force F = 8/x lb over the distance between x = 1 and x = 9 ft?

8 ln 9 ft-lb