Work-Energy Theorem

The Gleaners, Jean-François Millet, 1857

What was the definition of energy?�Energy: The ability to do work.

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The Work-Energy Theorem�The total work done on an object is equal to the change in its kinetic energy.

W = ΔK

F_netd = ½ mv² – ½ mv₀²

Example

Corey O’Graff exerts a force of 11.0 N at 29.0^o above the horizontal on a 6.40 kg sled for a distance of 2.00 m. The sled had an initial speed of 0.500 m/s and a coefficient of friction of 0.06.

1. What work is done by the boy?

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W_boy = F_boydcosθ

= (11.0 N)(2.00 m)(cos 29.0^o)

= 19.2 J

Example

Corey O’Graff exerts a force of 11.0 N at 29.0^o above the horizontal on a 6.40 kg sled for a distance of 2.00 m. The sled had an initial speed of 0.500 m/s and a coefficient of friction of 0.060.

b) What is the work done by friction?

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W_friction = F_frictiondcosθ

= μF_Ndcosθ

= (0.060)(6.40kg x 9.8m/s²)(2.00 m)(cos180^o)

= -7.5 J

Answers

1. W_boy = 19.2 J

Example

Corey O’Graff exerts a force of 11.0 N at 29.0^o above the horizontal on a 6.40 kg sled for a distance of 2.00 m. The sled had an initial speed of 0.500 m/s and a coefficient of friction of 0.060.

c) What is the total work done on the sled?

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W_total = ∑W

= 19.2 J + -7.5 J

= 11.7 J

Answers

1. W_boy = 19.2 J
2. W_friction = -7.5 J

Example

Corey O’Graff exerts a force of 11.0 N at 29.0^o above the horizontal on a 6.40 kg sled for a distance of 2.00 m. The sled had an initial speed of 0.500 m/s and a coefficient of friction of 0.060.

d) What is the final speed of the sled?

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W_total = ΔK

W_total = ½ mv² – ½ mv₀²

11.7 J = ½(6.40 kg)(v²) – ½(6.40kg)(0.500m/s)²

v = 1.98 m/s

Answers

1. W_boy = 19.2 J
2. W_friction = -7.5 J
3. W_total = 11.7 J

Example

To accelerate a certain car from rest to the speed v requires the work W₁. The work needed to accelerate the car from v to 2v is W₂. Which of the following is correct:

1. W₂ = W₁
2. W₂ = 2 W₁
3. W₂ = 3 W₁
4. W₂ = 4 W₁

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Step 1: Solve for W₁ in terms of v

W₁ = ½ mv² – ½ m(0)²

= ½ mv²

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Step 2: Solve for W₂ in terms of v

W₂ = ½ m(2v)² – ½ mv²

= ½ m(4v²) – ½ mv²

= 3/2 mv² = 3(½ mv²)

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Conclusion:

W₂ = 3 W₁

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Graphical Analysis of Work

If a 5 N net horizontal force is applied to an object moving it a horizontal distance of 4 m, the work done is:

W = Fd = 5 N x 4 m = 20 J

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Graphing force vs. position we get:

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What property of this graph would represent the work done on the object?

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position (m)

force (N)

Graphical Analysis of Work

Area!

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Area = length x width

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Area = F · Δx

= Fd

= W

= 20 J

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position (m)

force (N)

Example

A certain object experiences changing net forces that cause it to move a total of 20 m. A graph of the net force vs. position is shown to the right. What work is done on the object?

W = 10 J + 30 J + 50 J + 25 J

= 115 J

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position (m)

force (N)

10 J

30 J

50 J

25 J