How much work is done by a person lifting a 10lb10\operatorname{lb} bucket 4ft4\operatorname{ft} off the ground?

40 ft ⁡ ⋅ lb ⁡ s 40\operatorname{ft}\cdot\operatorname{lb}s

How much work is done by a person lifting a 10 lb 10\operatorname{lb} bucket 4 ft 4\operatorname{ft} off the ground?

How much work is done by a person lifting a 10 pound bucket four feet off the ground? So this is a work problem we're trying to solve, and it's actually pretty straightforward. Notice that we have that the bucket weighs 10 pounds. This is gonna be our force, the the weight force of the object. And I can see that the bucket here is not changing weight at any point. So because of that, we can say this is a constant force, meaning we can use the constant force equation for work. Work equals force times distance. The force is given to us as the weight of 10 pounds, and the distance we're lifting this is four feet off the ground. So we have 10 times four, which comes out to 40, and the units are going to be foot pounds for our work. So this right here is the solution to the problem, and that's how you can calculate the work here. Hope you found this video helpful.

How much work is done by a person lifting a 10 lb ⁡ 10\operatorname{lb} bucket 4 ft ⁡ 4\operatorname{ft} off the ground?

40 ft ⁡ ⋅ lb ⁡ s 40\operatorname{ft}\cdot\operatorname{lb}s

100 ft ⁡ ⋅ lb ⁡ s 100\operatorname{ft}\cdot\operatorname{lb}s

0 ft ⁡ ⋅ lb ⁡ s 0\operatorname{ft}\cdot\operatorname{lb}s

400 ft ⁡ ⋅ lb ⁡ s 400\operatorname{ft}\cdot\operatorname{lb}s

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
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7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
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- Area Between Curves
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10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
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- Introduction to Power Series
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16. Parametric Equations & Polar Coordinates7h 58m

10. Physics Applications of Integrals

Work

Calculus

10. Physics Applications of Integrals

Work

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Multiple Choice

How much work is done by a person lifting a $10 lb$ bucket $4 ft$ off the ground?

$100 ft times lb of s$

$0 ft times lb of s$

$40 ft times lb of s$

$400 ft times lb of s$

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Verified step by step guidance

Understand the problem: Work is defined as the product of force and displacement in the direction of the force. Here, the force is the weight of the bucket (10 lb), and the displacement is the vertical distance the bucket is lifted (4 ft).

Express the formula for work: Work = Force × Displacement. In this case, the force is constant because the weight of the bucket does not change as it is lifted.

Substitute the given values into the formula: Force = 10 lb and Displacement = 4 ft. Therefore, Work = 10 × 4.

Recognize the units: The work will be expressed in foot-pounds (ft⋅lb), which is the standard unit for work in this context.

Perform the multiplication to find the work done: Multiply the force (10 lb) by the displacement (4 ft) to calculate the total work. The result will be in ft⋅lb.