Energy and Work | Open Textbooks

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Work and Energy

Work is the transfer of energy when a force moves an object. Energy is the capacity to do work. These concepts are fundamental to understanding how physical systems change and interact.

Work

Work is done when a force causes displacement in the direction of the force.

Formula: W = Fd cos(θ)

Where: W = work (Joules, J), F = force (N), d = displacement (m), θ = angle between force and displacement

Key Points About Work:

Work is a scalar quantity (no direction, just magnitude)
Unit: Joule (J) = Newton × meter (N·m)
When force is parallel to displacement (θ = 0°): W = Fd
When force is perpendicular to displacement (θ = 90°): W = 0
Negative work occurs when the force opposes motion (θ > 90°)

Types of Energy

Energy Type	Formula	Description
Kinetic Energy (KE)	KE = ½mv²	Energy of motion; depends on mass and velocity squared
Gravitational PE	PE = mgh	Energy due to height above a reference point
Elastic PE	PE = ½kx²	Energy stored in a compressed/stretched spring
Mechanical Energy	ME = KE + PE	Total of kinetic and potential energy

Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy:

W_net = ΔKE = KE_f - KE_i = ½mv_f² - ½mv_i²

This theorem provides a powerful alternative to Newton's laws for solving many problems.

Conservation of Energy

Law of Conservation of Energy: Energy cannot be created or destroyed, only transformed from one form to another.

In a closed system with no non-conservative forces (like friction):

KE_i + PE_i = KE_f + PE_f

Types of Forces:

Conservative Forces	Non-Conservative Forces
Gravity	Friction
Spring force	Air resistance
Electrostatic force	Applied pushing/pulling
With non-conservative forces: KE_i + PE_i + W_nc = KE_f + PE_f

Power

Power is the rate at which work is done or energy is transferred.

P = W/t = Fv

Unit: Watt (W) = Joule/second (J/s)

1 horsepower (hp) ≈ 746 W

Energy Problem-Solving Strategy

Identify the system and what types of energy are involved
Choose your reference point for potential energy (usually ground = 0)
Determine if the system is conservative (no friction) or non-conservative
Write the energy equation: KE_i + PE_i + W_nc = KE_f + PE_f
Substitute known values and solve for the unknown

Common Energy Scenarios

Scenario	Energy Transformation
Ball falling	PE → KE (gravitational PE converts to kinetic)
Ball thrown upward	KE → PE (kinetic converts to gravitational PE)
Roller coaster	PE ↔ KE (continuous transformation)
Spring compressed then released	Elastic PE → KE
Car braking	KE → Heat (via friction—non-conservative)

💡 Examples

Work through these energy and work problems step by step.

Example 1: Calculating Work

Problem: A person pushes a 25 kg box across the floor with a force of 80 N for a distance of 5 m. The force is applied horizontally. How much work is done?

Solution

Given: F = 80 N, d = 5 m, θ = 0° (force parallel to displacement)

Work:

W = Fd cos(θ) = (80)(5)cos(0°) = (80)(5)(1) = 400 J

Example 2: Work at an Angle

Problem: A person pulls a sled with a force of 100 N at an angle of 30° above horizontal for 20 m. How much work is done?

Solution

Given: F = 100 N, d = 20 m, θ = 30°

Work:

W = Fd cos(θ) = (100)(20)cos(30°) = (100)(20)(0.866) = 1732 J

Example 3: Kinetic Energy

Problem: A 1200 kg car is traveling at 20 m/s. (a) What is its kinetic energy? (b) If it speeds up to 30 m/s, what is its new kinetic energy?

Solution

Given: m = 1200 kg, v₁ = 20 m/s, v₂ = 30 m/s

(a) At 20 m/s:

KE = ½mv² = ½(1200)(20)² = ½(1200)(400) = 240,000 J

(b) At 30 m/s:

KE = ½mv² = ½(1200)(30)² = ½(1200)(900) = 540,000 J

Note: Speed increased by 1.5×, but KE increased by 2.25× (because KE ∝ v²)

Example 4: Conservation of Energy

Problem: A 2 kg ball is dropped from a height of 10 m. What is its speed just before hitting the ground? (Ignore air resistance, use g = 10 m/s²)

Solution

Given: m = 2 kg, h = 10 m, v_i = 0 (dropped)

Using conservation of energy:

PE_i + KE_i = PE_f + KE_f

mgh + 0 = 0 + ½mv²

mgh = ½mv² (mass cancels)

gh = ½v²

v² = 2gh = 2(10)(10) = 200

v = √200 = 14.1 m/s

Example 5: Work-Energy with Friction

Problem: A 5 kg block slides down a 3 m ramp (height = 2 m) with friction. If it reaches the bottom at 4 m/s, how much energy was lost to friction?

Solution

Given: m = 5 kg, h = 2 m, v_f = 4 m/s, v_i = 0, g = 10 m/s²

Initial energy: E_i = PE_i = mgh = (5)(10)(2) = 100 J

Final energy: E_f = KE_f = ½mv² = ½(5)(4)² = ½(5)(16) = 40 J

Energy lost to friction:

W_friction = E_i - E_f = 100 - 40 = 60 J

✏️ Practice

Solve these work and energy problems. Use g = 10 m/s² where needed.

1. How much work is done when a 50 N force moves an object 8 m in the direction of the force?

2. A person lifts a 20 kg box to a shelf 1.5 m high. How much work is done against gravity?

3. A force of 60 N is applied at a 45° angle to pull a cart 10 m. How much work is done?

4. What is the kinetic energy of a 0.5 kg ball traveling at 12 m/s?

5. A 70 kg person climbs stairs to a height of 4 m. What is their gain in gravitational potential energy?

6. A roller coaster car (500 kg) is at rest at the top of a 30 m hill. What is its speed at the bottom? (Ignore friction)

7. A spring with k = 200 N/m is compressed 0.1 m. How much elastic potential energy is stored?

8. A 1000 W motor lifts an object. How much work does it do in 30 seconds?

9. A 2 kg ball is thrown upward with a speed of 15 m/s. How high does it rise before momentarily stopping?

10. A car (1500 kg) moving at 25 m/s brakes to a stop. How much work is done by the brakes?

Answer Key

W = Fd = (50)(8) = 400 J
W = mgh = (20)(10)(1.5) = 300 J
W = Fd cos(45°) = (60)(10)(0.707) = 424 J
KE = ½mv² = ½(0.5)(12)² = ½(0.5)(144) = 36 J
PE = mgh = (70)(10)(4) = 2800 J
mgh = ½mv² → gh = ½v² → v = √(2gh) = √(2×10×30) = √600 = 24.5 m/s
PE = ½kx² = ½(200)(0.1)² = ½(200)(0.01) = 1 J
P = W/t → W = Pt = (1000)(30) = 30,000 J
½mv² = mgh → h = v²/(2g) = (15)²/(20) = 225/20 = 11.25 m
W = ΔKE = 0 - ½mv² = -½(1500)(25)² = -½(1500)(625) = -468,750 J (negative because brakes oppose motion)

✅ Check Your Understanding

1. Can work be negative? When does this happen?

Show Answer

Yes, work can be negative. Negative work occurs when the force and displacement are in opposite directions (or when the angle between them is greater than 90°). Examples include: friction doing negative work on a sliding object, brakes doing negative work on a car, and gravity doing negative work when you lift an object upward.

2. Why does kinetic energy depend on velocity squared rather than just velocity?

Show Answer

The v² relationship comes from the work-energy theorem. When you apply a force to accelerate an object, the work done equals the change in kinetic energy. Mathematically, integrating F = ma with respect to displacement leads to ½mv². This has important real-world implications: doubling your speed quadruples the energy needed to stop, which is why high-speed car accidents are so much more dangerous.

3. Why doesn't a person carrying a heavy box do any work on the box while walking horizontally at constant speed?

Show Answer

The person applies an upward force to hold the box, but the displacement is horizontal. Since the force (vertical) is perpendicular to the displacement (horizontal), θ = 90°, and cos(90°) = 0. Therefore W = Fd cos(90°) = 0. No work is done on the box in the physics sense, even though the person feels tired. (The tiredness comes from muscle contractions, not physics work.)

4. A ball is thrown upward and returns to the starting point. If air resistance is present, how does the final speed compare to the initial speed?

Show Answer

The final speed is less than the initial speed. Air resistance is a non-conservative force that always opposes motion. On the way up, it does negative work (slowing the ball). On the way down, it also does negative work (opposing the downward motion). Energy is lost to heat due to air resistance during both phases, so the ball returns with less kinetic energy and therefore less speed.

🚀 Next Steps

Review any concepts that felt challenging
Move on to the next lesson when ready
Return to practice problems periodically for review