5.1 Work done and power

Achievable AP Physics 1

5. Work, energy, and power

Our AP Physics 1 course is currently in development and is a work-in-progress.

Work done and power

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In this subchapter, you’ll build the core ideas behind work, energy, and power. Here’s what we’ll cover:

Definition of work
Work done by common forces
Work and kinetic energy conversion
Power

Definition of work

When a constant force $F$ acts on a particle and produces a displacement $Δ r$ , the work $W$ done by the force is

$W = F \cdot Δ r = F Δ r cos θ,$

where $θ$ is the angle between $F$ and the displacement vector $Δ r$ .

A useful way to read this is: only the component of the force along the displacement contributes to work.

If $θ < 9 0^{\circ}$ , then $W > 0$ (force has a component along displacement).
If $θ > 9 0^{\circ}$ , then $W < 0$ (force opposes displacement).
If $θ = 9 0^{\circ}$ , then $W = 0$ (force is perpendicular to displacement).

Work by a variable force

If the force changes with position, you can’t use a single constant value of $F$ . Instead, you add up the work done over many tiny displacements.

For a force $F (x)$ varying along the $x$ -axis from $x = a$ to $x = b$ , the work is

$W = \int_{a}^{b} F (x) d x .$

Geometrically, this integral is the area under the $F$ vs. $x$ graph from $a$ to $b$ .

Work-kinetic energy theorem

Work is closely tied to changes in speed. For a particle of mass $m$ subject to a net force $F_{net}$ that changes its speed from $v_{1}$ to $v_{2}$ , the net work done equals the change in kinetic energy:

$W_{net} = \int_{r_{1}}^{r_{2}} F_{net} \cdot d r = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} .$

This is the work-energy theorem.

Work by common forces

Here are a few work results you’ll use often.

Gravity: A mass $m$ moves vertically by $Δ y$ :

$W_{g} = - m g Δ y .$

This sign convention matches the idea that gravity does negative work when an object moves upward ( $Δ y > 0$ ) and positive work when it moves downward ( $Δ y < 0$ ).
Spring: Hooke’s law $F = - k x$ yields

$W_{s} = \int_{x_{i}}^{x_{f}} (- k x) d x = - \frac{1}{2} k (x_{f}^{2} - x_{i}^{2}) .$
Normal and Tension: If displacement is along the surface or direction of the string, these can do work; otherwise zero.

Definition of power

Power tells you how quickly work is done.

For work $W$ over time $Δ t$ ,

$P = \frac{W}{Δ t} .$

Instantaneous power delivered by a force $F$ acting on a particle moving with velocity $v$ is

$P = \frac{d W}{d t} = F \cdot v .$

Example Problem 1

A block of mass $2 kg$ is pulled along a horizontal surface by a constant force $F = 10 N$ at an angle $3 0^{\circ}$ above the horizontal for a distance $5 m$ . Compute:
(a) The work done by $F$ .
(b) The work done by gravity and the normal force.
(c) The power if the block moves at constant speed $2 m/s$ .

Solution:

(spoiler)

The displacement is horizontal, so only the horizontal component of the force does work.

Horizontal component $F_{x} = F cos 3 0^{\circ} = 10 \times \frac{3}{2} = 53$ N, so

$W_{F} = F_{x} d = (53) (5) = 253 J .$
Gravity and the normal force are vertical while the motion is horizontal, so each is perpendicular to the displacement:

$\Rightarrow W_{g} = W_{N} = 0$ .
With constant speed, the instantaneous power from the pulling force is

$P = F \cdot v = F_{x} v = (53) (2) = 103 W .$

Example Problem 2

A car of mass $1000 kg$ accelerates from $20$ m/s to $30$ m/s on level ground in $5$ s. What is the average power output of the engine (neglect friction)?

Solution:

(spoiler)

The engine’s work goes into changing kinetic energy:

$Δ K = \frac{1}{2} m (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (3 0^{2} - 2 0^{2}) = 250, 000 J,$

so the average power is

$P_{avg} = \frac{Δ K}{Δ t} = \frac{250 , 000}{5} = 50, 000 W .$

Example Problem 3

A $3 kg$ mass attached to a spring ( $k = 200 N/m$ ) is stretched $0.1 m$ and released. How much work is done by the spring as it returns to equilibrium?

Solution:

(spoiler)

Use the spring‐work formula:

$W = \frac{1}{2} k x^{2} .$

Substitute $k = 200$ N/m and $x = 0.1$ m:

$W = \frac{1}{2} \times 200 \times (0.1)^{2} = 1.0 J .$

Hence, the spring does $1.0$ J of work returning the mass to equilibrium.

Spring mass system according to the question

Example Problem 4

A bicyclist applies a constant $100 N$ horizontally to move at $5 m/s$ . What power must the cyclist develop?

Solution:

(spoiler)

Because force and velocity point in the same direction,

$P = F \cdot v = F v = 100 \times 5 = 500 W .$

Example Problem 5

An object is acted upon by a variable force $F (x)$ whose dependence on the displacement $x$ is shown below. Calculate the work done by this force as the object moves from $x = 0$ to $x = 6$ m.

Solution:

(spoiler)

Because work is the area under the $F$ -vs.- $x$ graph, divide the shaded region into four simple shapes:

Triangle (0 to 1 m, height 3 N): $W_{1} = \frac{1}{2} (1) (3) = 1.5$ J.
Rectangle (1 to 3 m at 3 N): $W_{2} = 2 \times 3 = 6$ J.
Trapezoid (3 to 4 m, from 3 N to 1 N): $W_{3} = \frac{3 + 1}{2} \times 1 = 2$ J.
Rectangle (4 to 6 m at 1 N): $W_{4} = 2 \times 1 = 2$ J.

Total:

$W = W_{1} + W_{2} + W_{3} + W_{4} = 1.5 + 6 + 2 + 2 = 11.5 J .$

Definition of Work

Work: $W = F \cdot Δ r = F Δ r cos θ$
Only force component along displacement does work
Sign of work:
- $θ < 9 0^{\circ}$ : $W > 0$ (along displacement)
- $θ > 9 0^{\circ}$ : $W < 0$ (opposes displacement)
- $θ = 9 0^{\circ}$ : $W = 0$ (perpendicular)

Work by a Variable Force

For variable force: $W = \int_{a}^{b} F (x) d x$
Work equals area under $F$ vs. $x$ graph

Work-Kinetic Energy Theorem

Net work equals change in kinetic energy: $W_{net} = Δ K$
- $Δ K = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
Applies to net force acting over displacement

Work by Common Forces

Gravity: $W_{g} = - m g Δ y$
- Negative when moving up, positive when moving down
Spring (Hooke’s law): $W_{s} = - \frac{1}{2} k (x_{f}^{2} - x_{i}^{2})$
Normal and tension: Do work only if displacement has a component along their direction; otherwise zero

Definition of Power

Average power: $P = \frac{W}{Δ t}$
Instantaneous power: $P = F \cdot v$
Measures rate of doing work

Key Example Takeaways

Only force component parallel to displacement does work
Work by gravity and normal force is zero if displacement is perpendicular
For springs, use $W = \frac{1}{2} k x^{2}$ for work from stretched/compressed position to equilibrium
Power from constant force and velocity: $P = F v$ (if same direction)
For variable force, sum work by finding area under $F$ vs. $x$ graph

Core Toolkit Summary

Work: $W = F \cdot Δ r$
Variable force: integrate or use area under curve
Work-kinetic energy theorem: $W_{net} = Δ K$
Power: $P = \frac{W}{Δ t}$ and $P = F \cdot v$

Next | 5.2 Kinetic and potential energy

Work done and power

In this subchapter, you’ll build the core ideas behind work, energy, and power. Here’s what we’ll cover:

Definition of work
Work done by common forces
Work and kinetic energy conversion
Power

Definition of work

When a constant force $F$ acts on a particle and produces a displacement $Δ r$ , the work $W$ done by the force is

$W = F \cdot Δ r = F Δ r cos θ,$

where $θ$ is the angle between $F$ and the displacement vector $Δ r$ .

A useful way to read this is: only the component of the force along the displacement contributes to work.

If $θ < 9 0^{\circ}$ , then $W > 0$ (force has a component along displacement).
If $θ > 9 0^{\circ}$ , then $W < 0$ (force opposes displacement).
If $θ = 9 0^{\circ}$ , then $W = 0$ (force is perpendicular to displacement).

Work by a variable force

If the force changes with position, you can’t use a single constant value of $F$ . Instead, you add up the work done over many tiny displacements.

For a force $F (x)$ varying along the $x$ -axis from $x = a$ to $x = b$ , the work is

$W = \int_{a}^{b} F (x) d x .$

Geometrically, this integral is the area under the $F$ vs. $x$ graph from $a$ to $b$ .

Work-kinetic energy theorem

$W_{net} = \int_{r_{1}}^{r_{2}} F_{net} \cdot d r = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} .$

This is the work-energy theorem.

Work by common forces

Here are a few work results you’ll use often.

Gravity: A mass $m$ moves vertically by $Δ y$ :

$W_{g} = - m g Δ y .$

This sign convention matches the idea that gravity does negative work when an object moves upward ( $Δ y > 0$ ) and positive work when it moves downward ( $Δ y < 0$ ).
Spring: Hooke’s law $F = - k x$ yields

$W_{s} = \int_{x_{i}}^{x_{f}} (- k x) d x = - \frac{1}{2} k (x_{f}^{2} - x_{i}^{2}) .$
Normal and Tension: If displacement is along the surface or direction of the string, these can do work; otherwise zero.

Definition of power

Power tells you how quickly work is done.

For work $W$ over time $Δ t$ ,

$P = \frac{W}{Δ t} .$

Instantaneous power delivered by a force $F$ acting on a particle moving with velocity $v$ is

$P = \frac{d W}{d t} = F \cdot v .$

Example Problem 1

Solution:

(spoiler)

The displacement is horizontal, so only the horizontal component of the force does work.

Horizontal component $F_{x} = F cos 3 0^{\circ} = 10 \times \frac{3}{2} = 53$ N, so

$W_{F} = F_{x} d = (53) (5) = 253 J .$
Gravity and the normal force are vertical while the motion is horizontal, so each is perpendicular to the displacement:

$\Rightarrow W_{g} = W_{N} = 0$ .
With constant speed, the instantaneous power from the pulling force is

$P = F \cdot v = F_{x} v = (53) (2) = 103 W .$

Example Problem 2

A car of mass $1000 kg$ accelerates from $20$ m/s to $30$ m/s on level ground in $5$ s. What is the average power output of the engine (neglect friction)?

Solution:

(spoiler)

The engine’s work goes into changing kinetic energy:

$Δ K = \frac{1}{2} m (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (3 0^{2} - 2 0^{2}) = 250, 000 J,$

so the average power is

$P_{avg} = \frac{Δ K}{Δ t} = \frac{250 , 000}{5} = 50, 000 W .$

Example Problem 3

A $3 kg$ mass attached to a spring ( $k = 200 N/m$ ) is stretched $0.1 m$ and released. How much work is done by the spring as it returns to equilibrium?

Solution:

(spoiler)

Use the spring‐work formula:

$W = \frac{1}{2} k x^{2} .$

Substitute $k = 200$ N/m and $x = 0.1$ m:

$W = \frac{1}{2} \times 200 \times (0.1)^{2} = 1.0 J .$

Hence, the spring does $1.0$ J of work returning the mass to equilibrium.

Example Problem 4

A bicyclist applies a constant $100 N$ horizontally to move at $5 m/s$ . What power must the cyclist develop?

Solution:

(spoiler)

Because force and velocity point in the same direction,

$P = F \cdot v = F v = 100 \times 5 = 500 W .$

Example Problem 5

An object is acted upon by a variable force $F (x)$ whose dependence on the displacement $x$ is shown below. Calculate the work done by this force as the object moves from $x = 0$ to $x = 6$ m.

Solution:

(spoiler)

Because work is the area under the $F$ -vs.- $x$ graph, divide the shaded region into four simple shapes:

Triangle (0 to 1 m, height 3 N): $W_{1} = \frac{1}{2} (1) (3) = 1.5$ J.
Rectangle (1 to 3 m at 3 N): $W_{2} = 2 \times 3 = 6$ J.
Trapezoid (3 to 4 m, from 3 N to 1 N): $W_{3} = \frac{3 + 1}{2} \times 1 = 2$ J.
Rectangle (4 to 6 m at 1 N): $W_{4} = 2 \times 1 = 2$ J.

Total:

$W = W_{1} + W_{2} + W_{3} + W_{4} = 1.5 + 6 + 2 + 2 = 11.5 J .$

Achievable AP Physics 1

5. Work, energy, and power

Our AP Physics 1 course is currently in development and is a work-in-progress.