Work-Energy Theorem: Imagine pushing a heavy box across a room or lifting a backpack onto a shelf. These actions involve effort, motion, and energy, and the work-energy theorem helps us understand how they’re connected. This fundamental principle in physics states that the work done on an object equals the change in its kinetic energy—the energy it has due to motion. Whether you’re a student curious about how things move or an engineer designing a machine, the work-energy theorem is a powerful tool to analyze motion without getting bogged down by complex forces.
In this article, we’ll explore the work-energy theorem through its formula, derivation, and real-world examples. We’ll look at how it applies to scenarios with and without friction, constant and variable forces, and even visualize it with graphs. Plus, we’ll solve five problems to show how it works in action. By the end, you’ll see why this theorem is a cornerstone of physics, from everyday tasks to cutting-edge technology.
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What is the Work-Energy Theorem?
The work-energy theorem says that the work done on an object changes its kinetic energy, the energy an object has because it’s moving. Work is the transfer of energy when a force moves an object over a distance, like pushing a cart. Kinetic energy depends on an object’s mass and speed—think of a speeding car having more kinetic energy than a slow one.
For example, when you push a book across a table, the effort (work) you apply speeds it up, increasing its kinetic energy. The theorem simplifies physics problems by focusing on energy changes rather than calculating every force involved. It’s like a shortcut to understand motion, used in everything from sports (e.g., calculating a baseball’s speed after a hit) to engineering (e.g., designing efficient vehicles). By linking work and energy, the theorem helps us predict how objects behave when forces act on them.
Work-Energy Theorem Formula
The work-energy theorem is expressed as:
W = ΔK = K_f - K_i = ½mv_f² - ½mv_i²
Here’s what the variables mean:
- W: Work done on the object (in joules, J).
- ΔK: Change in kinetic energy (in joules, J).
- K_f: Final kinetic energy, ½mv_f².
- K_i: Initial kinetic energy, ½mv_i².
- m: Mass of the object (in kilograms, kg).
- v_f: Final velocity (in meters per second, m/s).
- v_i: Initial velocity (in meters per second, m/s).
Work is calculated as W = Fd cosθ, where F is force, d is displacement, and θ is the angle between the force and displacement directions. The unit for both work and kinetic energy is the joule (1 J = 1 kg·m²/s²), showing they’re two sides of the same coin—energy transfer.
| Variable | Symbol | Definition | Unit |
|---|---|---|---|
| Work | W | Force times displacement (cosine of angle) | Joule (J) |
| Kinetic Energy | K | Energy due to motion | Joule (J) |
| Mass | m | Object’s mass | Kilogram (kg) |
| Velocity | v | Speed with direction | Meter per second (m/s) |
Work-Energy Theorem Derivation
Let’s derive the work-energy theorem for a constant force to see why it works. Start with Newton’s second law: F = ma, where F is the net force, m is mass, and a is acceleration. Work is defined as W = Fd cosθ, where d is the displacement and θ is the angle between force and displacement.
Substitute F = ma into the work formula:
W = (ma)d cosθ
Assume the force is along the displacement (θ = 0, so cos 0 = 1), simplifying to W = mad. Now, use a kinematic equation to relate acceleration and displacement. The equation v_f² = v_i² + 2ad gives:
a = (v_f² - v_i²)/(2d)
Substitute a into the work equation:
W = m ((v_f² - v_i²)/(2d)) d = ½m v_f² - ½m v_i²
This is the change in kinetic energy (ΔK = K_f - K_i), so:
W = ΔK
For variable forces (e.g., a spring), work is calculated using calculus: W = ∫ F(x) dx. This integral sums the force over displacement, leading to the same result: work equals the change in kinetic energy. This derivation shows the theorem’s universal applicability.
Work-Energy Theorem Example
Consider pushing a 10 kg box across a frictionless floor with a constant force of 50 N over 5 meters. The force is applied horizontally (θ = 0). Calculate the box’s final velocity if it starts from rest (v_i = 0).
Work done is:
W = Fd cos 0 = 50 · 5 · 1 = 250 J
By the work-energy theorem, W = ΔK = ½mv_f² - ½mv_i². Since v_i = 0:
250 = ½ · 10 · v_f²
250 = 5 v_f²
v_f² = 50
v_f = √50 ≈ 7.07 m/s
The box reaches 7.07 m/s. This example shows how the theorem simplifies motion analysis by focusing on energy rather than forces and accelerations.
Work-Energy Theorem Graph
Graphically, the work-energy theorem can be visualized using a force vs. displacement graph. The area under the curve represents the work done. For a constant force, the graph is a horizontal line, and the area (force × displacement) equals W. For a variable force, like a spring (F = -kx), the graph is a line with a negative slope, and the area is a triangle or curve, calculated via integration.
A kinetic energy vs. time graph can show how kinetic energy changes as work is done. For example, if work increases kinetic energy linearly (constant force), the graph is a parabola since K = ½mv² and velocity increases with time. These graphs help visualize how work translates into motion, making the theorem intuitive.
Work-Energy Theorem with Friction
Friction complicates the work-energy theorem because it’s a non-conservative force, converting some work into heat rather than kinetic energy. The net work is:
W_net = W_applied - W_friction = ΔK
Suppose you push a 10 kg box with 50 N over 5 m, but friction exerts 20 N. The friction force opposes motion, so:
W_applied = 50 · 5 = 250 J
W_friction = 20 · 5 = 100 J
W_net = 250 - 100 = 150 J
If the box starts from rest:
150 = ½ · 10 · v_f²
v_f² = 30
v_f ≈ 5.48 m/s
Friction reduces the final velocity compared to the frictionless case (7.07 m/s).
| Scenario | Work Done | Effect on Kinetic Energy |
|---|---|---|
| No Friction | W = Fd cosθ | Full work increases ΔK |
| With Friction | W_net = (F - f)d cosθ | Reduced ΔK due to friction |
Work-Energy Theorem for Constant Force
For constant forces, like gravity or a steady push, the theorem applies directly. Consider lifting a 5 kg object 2 m vertically against gravity (g = 9.8 m/s²). The force is F = mg = 5 · 9.8 = 49 N. Work done is:
W = Fd cos 0 = 49 · 2 = 98 J
If the object starts at rest and isn’t moving at the top, ΔK = 0, meaning the work increases potential energy instead. For horizontal motion, like the box example, the work directly increases kinetic energy.
Work-Energy Theorem for Variable Force
Variable forces, like a spring’s force (F = -kx), require calculus. Work is:
W = ∫ F(x) dx
For a spring stretched from x = 0 to x = 0.1 m with spring constant k = 200 N/m:
W = ∫_0^0.1 (-kx) dx = -½kx² |_0^0.1 = -½ · 200 · (0.1)² = -1 J
The negative sign indicates work done against the spring, storing potential energy. This work changes the object’s kinetic energy if it’s released.
Work-Energy Theorem Problems with Solutions
1. Box, No Friction
A 10 kg box is pushed with 40 N over 4 m (frictionless, horizontal). Find final velocity (v_i = 0).
Solution:
W = 40 · 4 = 160 J
160 = ½ · 10 · v_f²
v_f² = 32, v_f ≈ 5.66 m/s.
Answer: 5.66 m/s.
2. Box with Friction
Same box, but friction is 10 N. Find final velocity.
Solution:
W_net = (40 - 10) · 4 = 120 J
120 = ½ · 10 · v_f²
v_f² = 24, v_f ≈ 4.90 m/s.
Answer: 4.90 m/s.
3. Lifted Object
A 2 kg object is lifted 3 m vertically (g = 9.8 m/s²). Find work done.
Solution:
F = 2 · 9.8 = 19.6 N
W = 19.6 · 3 = 58.8 J
If no velocity change, ΔK = 0, work increases potential energy.
Answer: 58.8 J.
4. Spring Stretched
A spring (k = 100 N/m) is stretched 0.2 m. Find work done.
Solution:
W = ∫_0^0.2 (-100x) dx = -½ · 100 · (0.2)² = -2 J
Answer: 2 J (work done on spring).
5. Projectile with Air Resistance
A 1 kg projectile (v_i = 10 m/s) slows to 8 m/s due to air resistance over 5 m. Find net work.
Solution:
ΔK = ½ · 1 · (8² - 10²) = ½ · (64 - 100) = -18 J
W_net = -18 J (negative due to resistive force).
Answer: -18 J.
| Problem | Scenario | Knowns | Solution | Answer |
|---|---|---|---|---|
| 1 | Box, no friction | F = 40 N, d = 4 m, m = 10 kg | W = ΔK | 5.66 m/s |
| 2 | Box with friction | F = 40 N, f = 10 N, d = 4 m, m = 10 kg | W_net = ΔK | 4.90 m/s |
| 3 | Lifted object | m = 2 kg, h = 3 m, g = 9.8 m/s² | W = Fd | 58.8 J |
| 4 | Spring | k = 100 N/m, x = 0.2 m | W = ∫ -kx dx | 2 J |
| 5 | Projectile | m = 1 kg, v_i = 10 m/s, v_f = 8 m/s | W = ΔK | -18 J |
Work-Energy Theorem: Conclusion
The work-energy theorem is a powerful bridge between work and kinetic energy, simplifying how we analyze motion. Whether pushing a box, stretching a spring, or launching a projectile, it shows how energy transfers drive the world around us. Its versatility shines in both simple scenarios (no friction) and complex ones (variable forces or friction).
