Work and Kinetic Energy
Definition of work, kinetic energy, and the work-energy theorem
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⚡ Work and Kinetic Energy
What is Work?
In physics, work has a very specific meaning: work is done when a force causes a displacement.
Definition of Work
where:
- = work (J, joules)
- = magnitude of force (N)
- = magnitude of displacement (m)
- = angle between force and displacement vectors
Units: 1 joule (J) = 1 N·m = 1 kg·m²/s²
Key Points
- Force must cause displacement - No displacement = no work
- Only the component of force parallel to displacement does work
- Work can be positive, negative, or zero
💡 Important: Work is a scalar quantity (not a vector), even though it's calculated from two vectors.
Sign of Work
Positive Work ()
- Force has component in direction of motion ()
- Energy is added to the system
- Example: Pushing a box forward while it moves forward
Negative Work ()
- Force has component opposite to motion ()
- Energy is removed from the system
- Example: Friction on a sliding box (friction opposes motion)
Zero Work ()
- Force perpendicular to motion ()
- OR no displacement ()
- Example: Carrying a box horizontally (weight is perpendicular to motion)
Special Cases
Case 1: Force Parallel to Displacement ()
Maximum positive work.
Case 2: Force Opposite to Displacement ()
Maximum negative work.
Case 3: Force Perpendicular to Displacement ()
No work done.
Kinetic Energy
Kinetic energy is energy of motion:
where:
- = kinetic energy (J)
- = mass (kg)
- = speed (m/s)
Properties of Kinetic Energy
- Always positive (or zero) - is always positive
- Scalar quantity - no direction
- Depends on speed squared - Double the speed = 4× the kinetic energy
- Frame-dependent - KE depends on reference frame
The Work-Energy Theorem
The work-energy theorem connects work and kinetic energy:
Interpretation
- Net work done on an object equals the change in its kinetic energy
- If : object speeds up (gains KE)
- If : object slows down (loses KE)
- If : speed unchanged (constant KE)
Calculating Work by Multiple Forces
When multiple forces act on an object:
Method 1: Sum the work done by each force
Each force can do positive, negative, or zero work.
Method 2: Find net force first
where is angle between and .
Both methods give the same result!
Variable Forces and Work
For a variable force (force changes during motion):
The work equals the area under the force vs. position graph.
For AP Physics 1, you usually just need to recognize this graphically.
⚠️ Common Misconceptions
Misconception 1: Confusing Force and Work
❌ Wrong: "I'm doing work just by holding this box" ✅ Right: Holding stationary = no displacement = no work (even though it's tiring!)
Misconception 2: Forgetting the Angle
❌ Wrong: always ✅ Right: - angle matters!
Misconception 3: Thinking Work is a Vector
❌ Wrong: Work has direction ✅ Right: Work is a scalar (can be positive or negative, but that's not direction)
Misconception 4: Normal Force Always Does Zero Work
Usually true (when surface is level), but NOT always! On an incline or accelerating elevator, normal force can do work.
Problem-Solving Strategy
- Identify all forces acting on the object
- Draw a free body diagram
- For each force, determine:
- Magnitude
- Angle relative to displacement
- Sign of work (positive/negative/zero)
- Calculate work for each force:
- Sum to find net work
- Apply work-energy theorem if finding speed change
Real-World Applications
Stopping Distance
When brakes are applied, friction does negative work to remove kinetic energy:
Doubling speed requires 4 times the stopping distance!
Accelerating Vehicles
Engine does positive work to increase kinetic energy:
Sports
- Baseball: Work done by bat on ball increases ball's KE
- Braking in cars: Friction converts KE to thermal energy
Key Formulas Summary
| Concept | Formula | Units | |---------|---------|-------| | Work | | J (joules) | | Kinetic Energy | | J | | Work-Energy Theorem | | J | | Power (rate of work) | | W (watts) |
📚 Practice Problems
1Problem 1easy
❓ Question:
A 5 kg box is pushed 10 m across a floor by a horizontal force of 20 N. If the box starts from rest, what is its final speed? (Assume no friction.)
💡 Show Solution
Given Information:
- Mass: kg
- Displacement: m
- Applied force: N (horizontal)
- Initial velocity: m/s (starts from rest)
- No friction
Find: Final speed
Step 1: Calculate work done by applied force
Since force is horizontal and displacement is horizontal, :
Step 2: Apply work-energy theorem
Since there's no friction, J
Answer: The final speed is approximately 8.94 m/s or m/s.
2Problem 2medium
❓ Question:
A 2 kg block slides 5 m down a frictionless incline that makes a 30° angle with the horizontal. What is the work done by gravity?
💡 Show Solution
Given Information:
- Mass: kg
- Displacement along incline: m
- Angle of incline:
- Frictionless (no friction)
Find: Work done by gravity
Step 1: Identify the gravitational force
This force points vertically downward.
Step 2: Find the angle between force and displacement
- Gravity points straight down (vertical)
- Displacement is along the incline (30° below horizontal)
- Angle between them:
Step 3: Calculate work
Alternative Method: Use vertical displacement
Vertical drop: m
Work by gravity = J ✓
Answer: Gravity does 49 J of work (positive because it has a component along the motion).
Note: Even though the block moves 5 m along the incline, only the vertical component of that displacement matters for gravitational work!
3Problem 3hard
❓ Question:
A 1200 kg car traveling at 25 m/s applies its brakes and comes to a stop in 50 m. (a) What is the work done by friction? (b) What is the magnitude of the average frictional force?
💡 Show Solution
Given Information:
- Mass: kg
- Initial velocity: m/s
- Final velocity: m/s (stops)
- Displacement: m
(a) Find work done by friction
Step 1: Calculate initial kinetic energy
Step 2: Calculate final kinetic energy
Step 3: Apply work-energy theorem
The negative sign indicates friction removes energy from the car.
(b) Find magnitude of frictional force
Step 4: Use work formula
Since friction opposes motion, :
Alternative approach:
Answers:
- (a) Work done by friction: -375,000 J (or -375 kJ)
- (b) Magnitude of friction force: 7,500 N
Interpretation: The friction force of 7,500 N acting over 50 m removes all the car's kinetic energy, bringing it to rest.