AP Physics 1- 3.2 Work- Study Notes- New Syllabus
AP Physics 1-3.2 Work – Study Notes
AP Physics 1-3.2 Work – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Work
- Work–Energy Theorem
- Work from a Force–Displacement Graph
Work
Work is the amount of energy transferred into or out of a system by a force exerted on that system over a distance.
Work is mathematically expressed as the dot product: \( W = \vec{F} \cdot \vec{d} = F d \cos \theta \)
Work is a scalar quantity, but it can be positive, negative, or zero depending on the direction of force relative to displacement.
Conservative vs. Nonconservative Forces
Conservative force:
- Work done is path-independent; depends only on initial and final positions.
- The work done by a conservative force (e.g., gravity, spring force) is associated with a change in potential energy.
- If the system returns to its initial configuration, the net work done by a conservative force = 0.
Nonconservative force:
- Work is path-dependent.
- Examples: friction, air resistance.
- Work done by these forces typically transforms mechanical energy into other forms (heat, sound, etc.).
Key Properties of Work
- Only the component of force parallel to displacement does work and changes the system’s total energy: \( W = F_{\parallel} d \).
- The perpendicular component of force can change the direction of motion but not the object’s kinetic energy.
- Units of work: Joule (J) → \( 1 \, \text{J} = 1 \, \text{N·m} \).
Example:
A \( 5 \, \text{kg} \) block is pulled up a rough incline (length \( 10 \, \text{m} \), angle \( 30^\circ \)). Find the work done by:
- (i) Gravity (a conservative force)
- (ii) Friction (a nonconservative force, assume \( f = 10 \, \text{N} \))
▶️Answer/Explanation
Step (1): Work done by gravity
Vertical height gained = \( h = 10 \sin 30^\circ = 5 \, \text{m} \).
Work by gravity = \( W_g = -m g h = -(5)(9.8)(5) = -245 \, \text{J} \).
Step (2): Work done by friction
Work by friction = \( W_f = -f d = -(10)(10) = -100 \, \text{J} \).
Final Answer:
- Gravity does \( -245 \, \text{J} \) (conservative, depends only on height).
- Friction does \( -100 \, \text{J} \) (nonconservative, depends on path length).
Work–Energy Theorem
The work–energy theorem states: The net work done on an object by all forces is equal to the change in its kinetic energy.
Relevant equation:
\( W_{net} = \Delta KE = KE_f – KE_i = \tfrac{1}{2} m v_f^2 – \tfrac{1}{2} m v_i^2 \)
- Work is a mechanism for transferring energy into or out of a system.
Key Points
- An external force can change the configuration of a system. Only the parallel component of the force contributes to changing the system’s kinetic energy: \( W = F_{\parallel} d \).
- If the center of mass and the point of application of the force move the same distance, the system can be modeled as a single object, and only its kinetic energy changes.
- For nonconservative forces like friction, the energy dissipated is typically: \( E_{dissipated} = f \cdot d \), where \( f \) is the frictional force and \( d \) is the distance traveled.
Implications
- If net work is positive → kinetic energy increases → the object speeds up.
- If net work is negative → kinetic energy decreases → the object slows down.
- If net work = 0 → kinetic energy remains constant → the object moves at constant speed.
Example:
A \( 2 \, \text{kg} \) block is pushed with a horizontal force of \( 15 \, \text{N} \) across a rough surface for \( 4 \, \text{m} \). Friction force = \( 5 \, \text{N} \). The block starts from rest. Find its final speed using the work–energy theorem.
▶️Answer/Explanation
Step (1): Work by applied force → \( W_{app} = 15 \times 4 = 60 \, \text{J} \).
Step (2): Work by friction → \( W_f = -5 \times 4 = -20 \, \text{J} \).
Step (3): Net work → \( W_{net} = 60 – 20 = 40 \, \text{J} \).
Step (4): Apply Work–Energy Theorem → \( W_{net} = \Delta KE = \tfrac{1}{2} m v^2 – 0 \).
\( 40 = \tfrac{1}{2} (2) v^2 \) → \( 40 = v^2 \).
\( v = \sqrt{40} \approx 6.32 \, \text{m/s} \).
Final Answer: The block’s final speed = 6.3 m/s.
Work from a Force–Displacement Graph
Work can also be determined graphically. The work done by a force is equal to the area under the curve of a graph of force (\( F \)) as a function of displacement (\( x \)).
- If the force is constant: \( W = F \cdot d \) → the graph is a rectangle, and area = base × height.
- If the force varies with displacement: \( W = \int F(x) \, dx \) → work is the definite integral of the force function with respect to displacement.
Positive area (above the axis) corresponds to positive work, while negative area (below the axis) corresponds to negative work.
Example:
A force increases linearly from \( 0 \, \text{N} \) to \( 10 \, \text{N} \) as an object moves from \( x = 0 \) to \( x = 5 \, \text{m} \). Find the work done.
▶️Answer/Explanation
Step (1): The graph is a straight line (a triangle) with base \( d = 5 \, \text{m} \) and height \( F = 10 \, \text{N} \).
Step (2): Work = area under graph = area of triangle = \( \tfrac{1}{2} \times \text{base} \times \text{height} \).
\( W = \tfrac{1}{2} (5)(10) = 25 \, \text{J} \).
Final Answer: The work done is 25 J.