Q: A 50-pound object is lifted vertically 4 feet. What is the work done by the lifting force? (Use the definition of work for a constant force.)

$$W = 50 \times 4 = 200$$ foot-pounds

Q: A force of 30 newtons compresses a spring 0.3 meters. What is the spring constant $$k$$ according to Hooke's Law?

$$k = \frac{30}{0.3} = 100$$ \text{ N/m}$

Q: If 1 foot-pound (ft-lb) is approximately 1.35582 joules (J), how many joules are in 100 ft-lb?

$$100 \times 1.35582 = 135.582 J$$

Question 1

A 50-pound object is lifted vertically 4 feet. What is the work done by the lifting force? (Use the definition of work for a constant force.)

Accepted Answer

$$W = 50 	imes 4 = 200$$ foot-pounds

Question 2

Which of the following best describes the formula for work done by a variable force $$F(x)$$ moving an object from $$x = a to$$ $$x = b$$?

Accepted Answer

$$W = \int_a^b F(x) \, dx$$

Question 3

A force of 30 newtons compresses a spring 0.3 meters. What is the spring constant $$k$$ according to Hooke's Law?

Accepted Answer

$$k = \frac{30}{0.3} = 100$$ 	ext{ N/m}$

Question 4

Which of the following is the correct unit for work in the International System of Units (SI)?

Accepted Answer

joule (J)

Question 5

A force $$F(x) = 100x N $$compresses a spring from $$x = 0.3 m to$$ $$x = 0.6 m. $$What is the work done in this process?

Accepted Answer

$$W = \int_{0.3}^{0.6} 100x \, dx = 100 \left[ \frac{x^2$}{2} ight]_{0.3}^{0.6} = 100 \left( \frac{0.36}{2} - \frac{0.09}{2} ight) = 100 	imes 0.135 = 13.5 J$$

Question 6

Which law states that the force required to compress or stretch a spring is proportional to the distance from its natural length?

Accepted Answer

Hooke's Law

Question 7

A spherical tank of radius 8 ft is half full of oil weighing 50 lb/ft$$^3. $$What is the correct integral to compute the work required to pump all the oil out through a hole at the top?

Accepted Answer

$$W = \$$int_0$$^8 50\pi (256y - 32y^2$ + y^3$) \, dy$$

Question 8

Which of the following is the correct expression for the work done by a constant force $$F$$ over a distance $$D$$?

Accepted Answer

$$W = F D$$

Question 9

Newton’s Law of Universal Gravitation states that the force between two masses $$m_1$$ and $$m_2$$ separated by distance $$d is$$:

Accepted Answer

$$F = G \frac{m_1$ m_2$}{d^2$}$$

Question 10

If 1 foot-pound (ft-lb) is approximately 1.35582 joules (J), how many joules are in 100 ft-lb?

Accepted Answer

$$100 	imes 1.35582 = 135.582 J$$

Question 11

A force $$F(x)$$ moves an object along a straight line from $$x = 2 to$$ $$x = 5. $$Which integral gives the total work done?

Accepted Answer

$$W = \$$int_2$$^5 F(x) \, dx$$

Question 12

Which of the following is a real-world example where the work done by a variable force must be calculated using integration?

Accepted Answer

Pumping oil out of a spherical tank

Question 13

Coulomb’s Law describes the force between two charges $$q_1$$ and $$q_2$$ separated by distance $$d as$$:

Accepted Answer

$$F = k \frac{q_1$ q_2$}{d^2$}$$

Question 14

A spring with $$k = 80 N/m is $$stretched from its natural length by 0.5 m. What is the work done?

Accepted Answer

$$W = \$$int_0$$^{0.5} 80x \, dx = 80 \left[ \frac{x^2$}{2} ight]_0$$^{0.5}$$ = 80 	imes 0.125 = 10 J$$

Question 15

Which of the following is NOT a fundamental type of force discussed in the study materials?

Accepted Answer

Frictional force

Question 16

If a force $$F(x) = 5x^2$ N moves an object from x = 0$ to x = 2$ m, what is the work done?

Accepted Answer

$$W = \$$int_0$$^2 5x^2$ \, dx = 5 \left[ \frac{x^3$}{3} ight]_0^2$ = 5 	imes \frac{8}{3} = \frac{40}{3} J$$

Question 17

Which of the following best explains why integration is needed to calculate work done by a variable force?

Accepted Answer

Because the force changes as the object moves, so the total work is the sum of infinitesimal contributions.

Applications of Integration: Work Done by a Force

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