AP Physics C Mechanics- 3.2 Work- Study Notes- New Syllabus
AP Physics C Mechanics- 3.2 Work – Study Notes
AP Physics C Mechanics- 3.2 Work – Study Notes – per latest Syllabus.
Key Concepts:
- Work and the Nature of Forces: Conservative and Nonconservative
- Work as a Scalar Quantity
- Work Done by a Variable Force
- The Work–Energy Theorem
- Graphical Interpretation of Work
Work and the Nature of Forces: Conservative and Nonconservative
Work is the amount of energy transferred into or out of a system when a force acts on that system over a displacement.
\( \mathrm{W = \vec{F} \cdot \vec{d} = Fd \cos\theta} \)
- \( \mathrm{W} \): work done (joules, J)
- \( \mathrm{\vec{F}} \): applied force
- \( \mathrm{\vec{d}} \): displacement of the object
- \( \mathrm{\theta} \): angle between the force and displacement
Work is a scalar quantity. It is positive when energy is transferred to the system (force acts in the direction of motion), and negative when energy is transferred from the system.
Conservative Forces:
The work done by a conservative force depends only on the initial and final positions of the object, not on the path taken.
- Examples include gravitational, elastic (spring), and electrostatic forces.
Work done by a conservative force can be expressed as the negative change in potential energy:
\( \mathrm{W_c = -\Delta U} \)
If the system returns to its initial configuration, \( \mathrm{\Delta U = 0} \), so the total work done by a conservative force is zero:
\( \mathrm{W_c = 0} \text{ for a closed path} \)
- This means energy is fully recoverable — no energy is lost as heat or dissipation.
2. Potential Energy and Conservative Forces:
Potential energy is associated only with conservative forces.
For example:
- Gravitational potential energy: \( \mathrm{U_g = mgh} \)
- Elastic potential energy (spring): \( \mathrm{U_s = \dfrac{1}{2}kx^2} \)
These forms of stored energy depend only on the system’s configuration, not the motion’s history.
Nonconservative Forces:
A nonconservative force is one for which the work done depends on the path taken.
- Common examples: friction, air resistance, tension, and applied forces.
- Work done by these forces typically transforms mechanical energy into other forms, such as heat or sound, which cannot be fully recovered as mechanical energy.
\( \mathrm{W_{nc} \neq 0 \text{ for a closed path}} \)
Example:
A \( \mathrm{2.0 \, kg} \) object is lifted vertically upward through a height of \( \mathrm{5.0 \, m} \) at a constant speed. Calculate:
- (a) The work done by the applied force
- (b) The work done by gravity
- (c) The net work done on the object
- (d) Discuss whether the forces involved are conservative or nonconservative.
▶️ Answer / Explanation
Given:
- Mass \( \mathrm{m = 2.0 \, kg} \)
- Height \( \mathrm{h = 5.0 \, m} \)
- Acceleration due to gravity \( \mathrm{g = 9.8 \, m/s^2} \)
- Constant speed → Net acceleration = 0 → Net work = 0
(a) Work done by the applied force:
To lift the object at constant speed, the applied force equals the weight: \( \mathrm{F = mg = (2.0)(9.8) = 19.6 \, N} \).
Work done: \( \mathrm{W_{app} = Fd\cos\theta = (19.6)(5.0)\cos(0^\circ) = 98 \, J} \)
(b) Work done by gravity:
The gravitational force acts downward, opposite to displacement:
\( \mathrm{W_g = F_g d \cos(180^\circ) = (19.6)(5.0)(-1) = -98 \, J} \)
(c) Net work done on the object:
\( \mathrm{W_{net} = W_{app} + W_g = 98 – 98 = 0} \)
Since there is no acceleration, the object’s kinetic energy does not change.
(d) Force classification:
- Gravity is a conservative force — its work depends only on the initial and final heights, not on the path.
- The applied force here is an external (nonconservative) force — its work depends on how the object is moved.
Work as a Scalar Quantity
Work is a scalar quantity that represents the transfer of energy between a system and its surroundings when a force acts over a displacement. Although it is derived from the dot product of two vectors — force and displacement — work itself has no direction; it can only be positive, negative, or zero.
\( \mathrm{W = \vec{F} \cdot \vec{d} = Fd \cos\theta} \)
- \( \mathrm{W} \): work done (joules, J)
- \( \mathrm{F} \): magnitude of the applied force (N)
- \( \mathrm{d} \): displacement of the object (m)
- \( \mathrm{\theta} \): angle between the force and displacement vectors
Nature of Work:
Positive Work:
When the force and displacement are in the same direction (\( \mathrm{0^\circ \leq \theta < 90^\circ} \)). Energy is transferred to the system. Example: pushing an object forward.
Negative Work:
When the force and displacement are in opposite directions (\( \mathrm{90^\circ < \theta \leq 180^\circ} \)). Energy is transferred from the system. Example: friction opposing motion.
Zero Work:
When the force is perpendicular to displacement (\( \mathrm{\theta = 90^\circ} \)) or when there is no displacement. Example: a satellite moving in a circular orbit under gravity, or a person holding a stationary object.
Key Characteristics:
- Work depends on both the magnitude and direction of force relative to displacement.
- Work can change an object’s kinetic energy or internal energy.
- As a scalar quantity, it can be added or subtracted directly (no vector components needed).
Example:
| Type of Work | Condition | Example Situation | Work Expression |
|---|---|---|---|
| Positive Work | \( \mathrm{0^\circ \leq \theta < 90^\circ} \) | A person pushes a box forward. | \( \mathrm{W = Fd\cos\theta > 0} \) |
| Negative Work | \( \mathrm{90^\circ < \theta \leq 180^\circ} \) | Friction acting opposite to motion. | \( \mathrm{W = Fd\cos\theta < 0} \) |
| Zero Work | \( \mathrm{\theta = 90^\circ} \text{ or } \mathrm{d = 0} \) | Car turning in a circle at constant speed. | \( \mathrm{W = 0} \) |
Example:
A \( \mathrm{10 \, kg} \) box is pulled across a horizontal floor by a \( \mathrm{50 \, N} \) force acting at an angle of \( \mathrm{30^\circ} \) above the horizontal. The box is displaced \( \mathrm{5.0 \, m} \) along the floor. The frictional force opposing motion is \( \mathrm{20 \, N} \). Calculate:
- (a) The work done by the applied force
- (b) The work done by friction
- (c) The work done by the normal force
- (d) The net work on the box
▶️ Answer / Explanation
Given:
- \( \mathrm{F = 50 \, N}, \, d = 5.0 \, m}, \, \theta = 30^\circ, \, f = 20 \, N} \)
(a) Work done by applied force:
\( \mathrm{W_{app} = Fd\cos\theta = (50)(5.0)\cos(30^\circ)} \)
\( \mathrm{W_{app} = 250(0.866) = 216.5 \, J} \)
→ Positive work, since the force has a component in the direction of displacement.
(b) Work done by friction:
\( \mathrm{W_f = -f d = -(20)(5.0) = -100 \, J} \)
→ Negative work, since friction opposes motion.
(c) Work done by normal force:
The normal force acts vertically while displacement is horizontal:
\( \mathrm{\theta = 90^\circ \Rightarrow W_N = F_N d\cos(90^\circ) = 0} \)
→ Zero work, since the force is perpendicular to motion.
(d) Net work on the box:
\( \mathrm{W_{net} = W_{app} + W_f + W_N = 216.5 – 100 + 0 = 116.5 \, J} \)
Work Done by a Variable Force
When a force acting on an object changes in magnitude or direction along the path of motion, it is called a variable force. The total work done by such a force between two points is calculated using integration summing up infinitesimal contributions of work over the displacement path.
\( \mathrm{W = \displaystyle \int_a^b \vec{F} \cdot d\vec{r}} \)
- \( \mathrm{W} \): total work done from point \( \mathrm{a} \) to \( \mathrm{b} \)
- \( \mathrm{\vec{F}} \): variable force acting on the object
- \( \mathrm{d\vec{r}} \): infinitesimal displacement vector
- The dot product ensures that only the component of force along displacement contributes to the work.
Dot Product and Work:
The dot product of two vectors \( \mathrm{\vec{A}} \) and \( \mathrm{\vec{B}} \) gives a scalar quantity defined by:
\( \mathrm{\vec{A} \cdot \vec{B} = AB \cos\theta} \)
- \( \mathrm{\theta} \): angle between \( \mathrm{\vec{A}} \) and \( \mathrm{\vec{B}} \)
- The dot product measures how much of one vector lies in the direction of the other.
- Work, being the dot product of \( \mathrm{\vec{F}} \) and \( \mathrm{\vec{d}} \), is therefore a scalar quantity.
Work from the Parallel Component of Force:
Only the component of the force parallel to the displacement contributes to the work done on the system:
\( \mathrm{W = F_\parallel d = Fd\cos\theta} \)
- \( \mathrm{F_\parallel = F\cos\theta} \): component of the force along the displacement direction.
- If the force is completely perpendicular (\( \mathrm{\theta = 90^\circ} \)), no work is done.
- If the parallel component is constant, this formula directly applies to calculate total work.
Perpendicular Component of Force:
- A component of force perpendicular to the motion does no work it only changes the direction of the object’s velocity, not its magnitude.
- Examples: centripetal force in uniform circular motion, normal force on an object moving horizontally.
- Since these forces do not contribute to energy transfer, they do not change the system’s kinetic energy.
Work–Energy Connection:
- The net work done by all forces acting on an object equals the change in its kinetic energy.
- This principle remains valid for both constant and variable forces when the correct integral form is used.
Example:
A variable force \( \mathrm{F(x) = 2x} \, \text{N} \) acts on a particle along the x-axis, where \( \mathrm{x} \) is in meters. Find the work done by the force as the particle moves from \( \mathrm{x = 0 \, m} \) to \( \mathrm{x = 4 \, m} \).
▶️ Answer / Explanation
Given: \( \mathrm{F(x) = 2x} \), limits: \( \mathrm{x_1 = 0, \, x_2 = 4} \)
Work done is obtained using the integral:
\( \mathrm{W = \displaystyle \int_{0}^{4} F(x)\,dx = \int_{0}^{4} 2x \, dx} \)
Evaluating the integral:
\( \mathrm{W = [x^2]_0^4 = (4)^2 – 0^2 = 16 \, J} \)
Result: \( \mathrm{W = 16 \, J} \)
Conceptual Note: The force increases with displacement, so more work is done as the particle moves farther. This illustrates how integration accounts for the variable nature of force along a path.
The Work–Energy Theorem
The work–energy theorem states that the net work done by all forces acting on an object is equal to the change in its kinetic energy. This theorem connects the concept of work (energy transfer through force and displacement) with the change in motion (kinetic energy) of a system.
\( \mathrm{\Delta K = W_{net} = \sum_i W_i = \sum_i F_{i\parallel} d_i} \)
- \( \mathrm{\Delta K} \): change in kinetic energy
- \( \mathrm{W_{net}} \): total (net) work done by all forces
- \( \mathrm{F_{i\parallel}} \): component of the \( i^{th} \) force parallel to displacement
- \( \mathrm{d_i} \): displacement of the point of application of the force
Statement of the Theorem:
The change in an object’s kinetic energy equals the net work done on it:
\( \mathrm{W_{net} = \Delta K = K_f – K_i = \dfrac{1}{2}mv_f^2 – \dfrac{1}{2}mv_i^2} \)
- If \( \mathrm{W_{net} > 0} \), the object speeds up (kinetic energy increases).
- If \( \mathrm{W_{net} < 0} \), the object slows down (kinetic energy decreases).
- If \( \mathrm{W_{net} = 0} \), the object’s kinetic energy remains constant.
External Forces and System Configuration:
- An external force can change the configuration or shape of a system.
- The component of this force parallel to the displacement of its point of application determines how the system’s energy changes.
- Mathematically, this can be represented as:
\( \mathrm{\Delta K_{system} = F_{ext, \parallel} d} \)
- Here, only the portion of the external force aligned with the direction of displacement performs work that alters kinetic energy.
Modeling Systems as Objects:
- If the center of mass and the point of application of the external force move through the same displacement, the system behaves like a single object.
- In this case, only the kinetic energy of the system changes, and internal configuration remains constant.
- This simplification is often used when analyzing rigid-body motion or translational systems.
Energy Dissipation by Nonconservative Forces:
- Forces such as friction or air resistance convert mechanical energy into nonrecoverable forms, such as heat.
- The amount of energy dissipated due to friction is given by:
\( \mathrm{E_{mech} = F_f d \cos\theta} \)
- Since friction opposes motion, \( \mathrm{\theta = 180^\circ} \), so \( \mathrm{E_{mech} = -F_f d} \).
- This represents a loss in mechanical energy of the system.
Example:
A \( \mathrm{4.0 \, kg} \) object is pushed across a horizontal surface by a constant horizontal force of \( \mathrm{20 \, N} \). The surface exerts a frictional force of \( \mathrm{8.0 \, N} \). The object moves a distance of \( \mathrm{5.0 \, m} \) from rest. Find the change in kinetic energy of the object and its final speed.
▶️ Answer / Explanation
Given: \( \mathrm{F_{app} = 20 \, N}, \, F_f = 8 \, N, \, m = 4.0 \, kg, \, d = 5.0 \, m} \)
Step 1: Net work done on the object
\( \mathrm{W_{net} = (F_{app} – F_f)d = (20 – 8)(5.0) = 60 \, J} \)
Step 2: Apply the work–energy theorem
\( \mathrm{W_{net} = \Delta K = \dfrac{1}{2}mv_f^2 – \dfrac{1}{2}mv_i^2} \)
Since the object starts from rest, \( \mathrm{v_i = 0} \):
\( \mathrm{60 = \dfrac{1}{2}(4.0)v_f^2} \)
\( \mathrm{v_f^2 = 30 \Rightarrow v_f = 5.48 \, m/s} \)
Graphical Interpretation of Work
The work done by a force acting on an object can be represented graphically as the area under the curve of the component of force parallel to displacement, \( \mathrm{F_{\parallel}} \), plotted as a function of displacement.
\( \mathrm{W = \displaystyle \int F_{\parallel} \, dx} \)
- \( \mathrm{F_{\parallel}} \): component of the force along the direction of motion.
- \( \mathrm{dx} \): infinitesimal displacement along the motion.
- The total area under the \( \mathrm{F_{\parallel}} \) vs. \( \mathrm{x} \) curve gives the total work done between two positions.
Constant Force:
If \( \mathrm{F_{\parallel}} \) is constant, the graph of \( \mathrm{F_{\parallel}} \) vs. \( \mathrm{x} \) is a horizontal line.
- The area (and therefore work) is a rectangle:
\( \mathrm{W = F_{\parallel} \Delta x = Fd\cos\theta} \)
The work is proportional to both the constant force and the distance moved in the direction of the force.
Variable Force:
If the force varies with displacement, the graph of \( \mathrm{F_{\parallel}} \) vs. \( \mathrm{x} \) is not a straight line.
- In this case, the total work is obtained by integrating under the curve:
\( \mathrm{W = \displaystyle \int_{x_1}^{x_2} F_{\parallel}(x) \, dx} \)
- The area under the curve between two positions \( \mathrm{x_1} \) and \( \mathrm{x_2} \) represents the work done.
- If \( \mathrm{F_{\parallel}} \) is positive, work is positive (force aids motion); if negative, work is negative (force opposes motion).
Negative and Zero Work in Graphs:
- Regions below the x-axis represent negative work.
- Regions above the x-axis represent positive work.
- If equal positive and negative areas occur, the net work is zero.
Example:
The graph below shows \( \mathrm{F_{\parallel}} \) as a function of \( \mathrm{x} \) for a block being pushed along a horizontal surface. The force increases linearly from \( \mathrm{0 \, N} \) at \( \mathrm{x = 0 \, m} \) to \( \mathrm{6 \, N} \) at \( \mathrm{x = 3 \, m} \).
Find the work done by the force during this displacement.
▶️ Answer / Explanation
Given: The graph is a straight line from \( (0, 0) \) to \( (3, 6) \), forming a triangle under the \( \mathrm{F_{\parallel}} \)-axis.
Step 1: Area under the curve = work done.
Area = \( \mathrm{\dfrac{1}{2} \times base \times height = \dfrac{1}{2}(3)(6) = 9 \, J} \)
Step 2: Interpret the result.
- The force does \( \mathrm{9 \, J} \) of positive work on the block.
- Because the force increases with displacement, more work is done at later positions.
Final Answer: \( \mathrm{W = 9 \, J} \)