Physics

Chapter 7 Work and Kinetic Energy

Imron Rosyadi

Introduction

Chapter Outline

• 7.1 Work
• 7.2 Kinetic Energy
• 7.3 Work-Energy Theorem
• 7.4 Power

What is Work and Kinetic Energy?

• Work is done when a force causes displacement.
• Kinetic energy is the energy of motion.
• These concepts are fundamental to understanding how forces change an object’s motion.

7.1 Work

Learning Objectives

By the end of this section, you will be able to:

• Represent the work done by any force.
• Evaluate the work done for various forces.

Defining Work

• Work is the transfer of energy to an object via a force causing displacement.
• Infinitesimal work \(dW\) by a force \(\vec{F}\) through an infinitesimal displacement \(d\vec{r}\) is:

\[dW = \vec{F} \cdot d\vec{r} = |\vec{F}||d\vec{r}|\cos\theta\]

• Total work is the integral along the path:

\[W_{AB} = \int_{path\ AB} \vec{F} \cdot d\vec{r}\]

• Work is a scalar quantity, representing energy transferred.
• The dot product emphasizes that only the component of force parallel to displacement does work.
• The integral sums up tiny bits of work over the entire path.

Work by a Force: Visualizing

Figure 7.2: Vectors used to define work.

Key Aspects

• \(\vec{F}\) is the force vector.
• \(d\vec{r}\) is the infinitesimal displacement vector.
• \(\theta\) is the angle between \(\vec{F}\) and \(d\vec{r}\).
• Work can be positive, negative, or zero depending on \(\theta\).

Note

Units of Work:

In SI, work is measured in Joules (J), where \(1\ J = 1\ N \cdot m\).

In the English system, it’s foot-pounds (ft·lb).

• The image shows a curved path from A to B. At any point, the force vector and infinitesimal displacement vector are shown.
• Positive work: Force has a component in the direction of displacement (\(0^\circ \le \theta < 90^\circ\)).
• Negative work: Force has a component opposite to displacement (\(90^\circ < \theta \le 180^\circ\)).
• Zero work: Force is perpendicular to displacement (\(\theta = 90^\circ\)).
• Highlight the practical meaning of positive, negative, and zero work (e.g., pushing a box vs. friction vs. holding a briefcase).

Work by Constant Forces

When force \(\vec{F}\) is constant in magnitude and direction:

• The work simplifies to: \[W_{AB} = \vec{F} \cdot (\vec{r}_B - \vec{r}_A) = |\vec{F}||\vec{r}_B - \vec{r}_A|\cos\theta\]
• Or simply \(W = Fd\cos\theta\), where \(d\) is the total displacement.

Examples

• Pushing a lawn mower:
  • Force has a component in direction of motion.
  • Work done is positive.
• Holding a briefcase (no displacement):
  • Displacement is zero, so work done is zero.
• Walking horizontally with a briefcase:
  • Force (upward) is perpendicular to displacement (horizontal).
  • Work done is zero (\(\cos 90^\circ = 0\)).

Figure 7.3: Work done by a constant force.

• For a constant force, the integral reduces to a simple dot product of the force and the total displacement vector.
• Discuss each part of Figure 7.3:
  • 1. Lawn mower: Force has a horizontal component, so work is done. This is an example of positive work.
  • 2. Holding briefcase: Displacement is zero, so work is zero.
  • 3. Walking with briefcase: Force (gravity/normal) is vertical, displacement is horizontal, angle is 90 degrees, so work is zero. This highlights the importance of the angle.

Example 7.1: Pushing a Lawn Mower

Problem: A person pushes a lawn mower 25.0 m with a constant force of 75.0 N at a 35° angle below the horizontal. How much work is done?

Strategy:

Use \(W = Fd\cos\theta\).

Solution:

\(F = 75.0\ N\)

\(d = 25.0\ m\)

\(\theta = 35.0^\circ\)

\(W = (75.0\ N)(25.0\ m)\cos(35.0^\circ)\)

\(W = 1875\ N \cdot m \times 0.819\)

\(W = 1535.6\ J \approx 1.54 \times 10^3\ J\)

• This is a direct application of the constant force work formula.
• Emphasize unit consistency (N, m, J).
• Significance: Connects to real-world effort and energy scales.

Work by Contact Forces

Normal Force (\(N\))

• Always perpendicular to the surface.
• Displacement is tangent to the surface.
• Therefore, the normal force does no work in most simple cases. \(dW_N = \vec{N} \cdot d\vec{r} = 0\)

Friction Force (\(\vec{f}\))

• Opposite to the direction of motion for kinetic friction.
• Work done by kinetic friction is negative: \(W_{fr} = \int_{path\ AB} \vec{f}_k \cdot d\vec{r} = -f_k \int_{path\ AB} |d\vec{r}| = -f_k |\ell_{AB}|\) where \(|\ell_{AB}|\) is the path length.
• Static friction typically does no work as there is no relative displacement.

Figure 7.4: Top view of paths for moving a couch.

• Explain why normal force does no work: the angle is 90 degrees, \(\cos 90^\circ = 0\).
• Explain why kinetic friction does negative work: it opposes motion, so the angle is 180 degrees, \(\cos 180^\circ = -1\).
• Discuss Figure 7.4: two paths are shown, emphasizing that friction’s work depends on path length, not just displacement. This is a key characteristic of nonconservative forces.

Example 7.2: Moving a Couch

Problem: A couch (normal force 1 kN, \(\mu_K = 0.6\)) is pushed:

1. 3 m parallel to a wall, then 1 m perpendicular to the wall. How much work is done by friction?
2. Then moved straight back to its original position. What is the total work done against friction?

Strategy:

• \(f_K = \mu_K N\) (constant kinetic friction).
• Work by friction \(W_{fr} = -f_K d\), where \(d\) is path length.
• Work against friction is \(-W_{fr}\).

Solution:

1. \(f_K = (0.6)(1\ kN) = 0.6\ kN = 600\ N\).

   Path length \(d_a = 3\ m + 1\ m = 4\ m\).

   \(W_{fr,a} = -(600\ N)(4\ m) = -2400\ J = -2.4\ kJ\).

2. Return path (hypotenuse): \(d_b = \sqrt{(3\ m)^2 + (1\ m)^2} = \sqrt{10}\ m \approx 3.16\ m\).

   Total path length for round trip = \(4\ m + 3.16\ m = 7.16\ m\).

   Total work against friction = \((600\ N)(7.16\ m) = 4296\ J \approx 4.3\ kJ\).

• Emphasize that the path length is crucial for friction’s work.
• Work done by friction is negative, work against friction is positive.
• Significance: Even if the net displacement is zero (closed path), work by friction is non-zero, indicating it’s a nonconservative force.

Work by Gravity

Gravitational Force

• Near Earth’s surface, gravity (\(\vec{F}_g = -mg\hat{j}\)) is constant in magnitude (\(mg\)) and direction (vertically down).
• Work done by gravity over any path from A to B: \(W_{grav, AB} = -mg(y_B - y_A)\)

Key Points

• Depends only on the change in vertical height (\(y_B - y_A\)), not the path.
• Positive work if object moves downward (\(y_B < y_A\)).
• Negative work if object moves upward (\(y_B > y_A\)).
• Zero work if horizontal displacement.

Figure 7.5: Side view of the paths for moving a book to and from a shelf.

• Emphasize the coordinate system choice (y-axis positive upwards).
• Explain why gravity’s work depends only on vertical displacement: it’s a conservative force.
• Discuss Figure 7.5:
  • Moving down from shelf: positive work by gravity.
  • Moving horizontally: zero work by gravity.
  • Moving up to shelf: negative work by gravity.
  • Total work by gravity over a closed path is zero. This is the defining characteristic of a conservative force.

Work by Variable Forces

When forces vary in magnitude and/or direction:

• The infinitesimal work \(dW\) is: \(dW = F_x dx + F_y dy + F_z dz\)
• Total work is the line integral: \(W_{AB} = \int_{path\ AB} (F_x dx + F_y dy + F_z dz)\)

Example: Spring Force (Hooke’s Law)

• A perfectly elastic spring: \(\vec{F} = -k\Delta\vec{x}\) (where \(\Delta\vec{x}\) is displacement from equilibrium).
• Work done by spring when stretched/compressed from \(x_A\) to \(x_B\): \(W_{spring, AB} = -\frac{1}{2}k(x_B^2 - x_A^2)\)
• Work done against the spring is positive: \(\frac{1}{2}k(x_B^2 - x_A^2)\).

Figure 7.7: Spring force and displacement.

• The line integral is generally complex but simplifies if we can express one variable in terms of another (e.g., \(y\) in terms of \(x\)).
• Focus on the spring force as a common and important example of a variable force.
• Explain Hooke’s Law: \(k\) is the spring constant, \(\Delta x\) is the displacement from equilibrium. The negative sign means the force opposes the displacement.
• Derive/explain the work done by a spring: it’s the area under the force-displacement graph for the spring force (which is a straight line \(F=-kx\)).
• Highlight that spring force, like gravity, is a conservative force (work depends only on endpoints).

Work Done by a Spring Force

Problem: A spring requires 0.54 J of work to stretch 6 cm from equilibrium.

1. What is its spring constant \(k\)?

2. How much work is required to stretch it an additional 6 cm (total 12 cm)?

Strategy:

Work required = work done against the spring \(= \frac{1}{2}k(x_B^2 - x_A^2)\).

Solution:

1. For stretching from \(x_A=0\) to \(x_B=6\ cm\ (0.06\ m)\):

\(0.54\ J = \frac{1}{2}k(0.06\ m)^2\)

\(k = \frac{2 \times 0.54\ J}{(0.06\ m)^2} = \frac{1.08}{0.0036}\ N/m = 300\ N/m\)

2. For stretching from \(x_A=6\ cm\ (0.06\ m)\) to \(x_B=12\ cm\ (0.12\ m)\):

\(W = \frac{1}{2}(300\ N/m)[(0.12\ m)^2 - (0.06\ m)^2]\)

\(W = 150\ N/m[0.0144\ m^2 - 0.0036\ m^2]\)

\(W = 150\ N/m[0.0108\ m^2] = 1.62\ J\)

• Careful with units: convert cm to m.
• Reinforce that work against spring is positive as it’s energy stored.
• Significance: Work required is proportional to the square of displacement, so stretching twice as far requires four times the work from equilibrium. The work for the second 6cm stretch is not the same as the first 6cm stretch.

7.2 Kinetic Energy

Learning Objectives

By the end of this section, you will be able to:

• Calculate the kinetic energy of a particle given its mass and velocity or momentum.
• Evaluate the kinetic energy of a body relative to different frames of reference.

Definition of Kinetic Energy

• Kinetic energy (\(K\)) is the energy of motion.
• For a particle of mass \(m\) and speed \(v\): \[K = \frac{1}{2}mv^2\]
• For a system of particles, sum of individual kinetic energies.
• Can also be expressed in terms of momentum \(\vec{p} = m\vec{v}\): \[K = \frac{p^2}{2m}\]

Note

Units of Kinetic Energy:

Same as work: Joules (J).

\(1\ J = 1\ kg \cdot m^2/s^2\).

• Introduce kinetic energy as “energy of motion.”
• Emphasize that speed \(v\) is magnitude, so \(K\) is always positive or zero.
• Show the derivation from momentum to connect to previous concepts.
• Reinforce that the units are the same as work, hinting at the work-energy theorem.

Example 7.6: Kinetic Energy of an Object

Problem:

1. Kinetic energy of an 80-kg athlete running at 10 m/s?
2. Mass of an asteroid (traveling at 22 km/s) if its kinetic energy upon impact was \(4.2 \times 10^{23}\) J?
3. Kinetic energy of a thermal neutron (mass \(1.68 \times 10^{-27}\) kg) traveling at 2.2 km/s?

Strategy:

Use \(K = \frac{1}{2}mv^2\) or \(m = \frac{2K}{v^2}\). Convert units to SI.

Solution:

1. \(K = \frac{1}{2}(80\ kg)(10\ m/s)^2 = \frac{1}{2}(80)(100)\ J = 4000\ J = 4.0\ kJ\)

2. \(v = 22\ km/s = 22000\ m/s\)

\(m = \frac{2(4.2 \times 10^{23}\ J)}{(22000\ m/s)^2} = \frac{8.4 \times 10^{23}}{4.84 \times 10^8}\ kg \approx 1.7 \times 10^{15}\ kg\)

3. \(v = 2.2\ km/s = 2200\ m/s\)

\(K = \frac{1}{2}(1.68 \times 10^{-27}\ kg)(2200\ m/s)^2 = \frac{1}{2}(1.68 \times 10^{-27})(4.84 \times 10^6)\ J \approx 4.1 \times 10^{-21}\ J\)

• Highlight the wide range of kinetic energy values in nature.
• Discuss the units and conversions (km/s to m/s, kJ).
• Mention context for the asteroid energy (megatons of TNT) and neutron energy (electron-volts) to give students a sense of scale.

Kinetic Energy and Reference Frames

• Velocity is a relative quantity, so kinetic energy also depends on the frame of reference.
• Choose a frame that simplifies the analysis.

Example: Person in a Subway Car

• Person’s mass: 75.0 kg
• Speed relative to car (\(v_{PC}\)): 1.50 m/s
• Car speed relative to tracks (\(v_{CT}\)): 15.0 m/s

1. Relative to the car:

   \(K = \frac{1}{2}(75.0\ kg)(1.50\ m/s)^2 = 84.4\ J\)

2. Relative to the tracks:

   • If walking towards front: \(v_{PT} = v_{CT} + v_{PC} = 15.0 + 1.50 = 16.5\ m/s\) \(K = \frac{1}{2}(75.0\ kg)(16.5\ m/s)^2 = 10.2\ kJ\)
   • If walking towards back: \(v_{PT} = v_{CT} - v_{PC} = 15.0 - 1.50 = 13.5\ m/s\) \(K = \frac{1}{2}(75.0\ kg)(13.5\ m/s)^2 = 6.83\ kJ\)

3. Relative to a frame moving with the person:

   \(v = 0\ m/s\), so \(K = 0\ J\).

Figure 7.10: Person walking in a train car.

• Explain the vector addition of velocities (\(v_{PT} = v_{PC} + v_{CT}\)).
• Emphasize that kinetic energy is always non-negative.
• Discuss different types of kinetic energy: translational, rotational, vibrational, internal, thermal (briefly, as an introduction to later chapters).
• Reinforce that all forms are fundamentally the same physical quantity related to motion.

7.3 Work-Energy Theorem

Learning Objectives

By the end of this section, you will be able to:

• Apply the work-energy theorem to find information about a particle’s motion.
• Use the work-energy theorem to find information about forces acting on a particle.

The Work-Energy Theorem

• The net work done on a particle equals the change in the particle’s kinetic energy.

\[W_{net} = K_B - K_A = \Delta K\]

• Where:
  • \(W_{net}\) is the sum of work done by all forces.
  • \(K_A\) is the initial kinetic energy.
  • \(K_B\) is the final kinetic energy.

• This is a fundamental principle connecting work and kinetic energy.
• Show the derivation from Newton’s second law: \(dW_{net} = \vec{F}_{net} \cdot d\vec{r} = m \frac{d\vec{v}}{dt} \cdot d\vec{r} = m \frac{d\vec{r}}{dt} \cdot d\vec{v} = m\vec{v} \cdot d\vec{v}\). Integrating gives \(\frac{1}{2}m v_B^2 - \frac{1}{2}m v_A^2\).
• Emphasize the net work. Misidentifying forces is a common mistake.
• The theorem simplifies calculations in many cases where direct application of Newton’s second law is difficult (e.g., curved paths, variable forces).

Work-Energy Theorem: Illustration

Figure 7.11: Horse pulls a loaded cart.

Implications

• If \(W_{net} > 0\), kinetic energy increases (object speeds up).
• If \(W_{net} < 0\), kinetic energy decreases (object slows down).
• If \(W_{net} = 0\), kinetic energy is constant (speed remains the same).

Tip

Advantage: The work-energy theorem often avoids complex calculations involving acceleration and time, especially for variable forces or curved paths.

• Use Figure 7.11 to illustrate: the horses do positive work, increasing the cart’s kinetic energy.
• Explain why the work-energy theorem is powerful: it’s a scalar equation involving only speeds and positions, rather than vector equations of forces and accelerations.
• Give an example of sliding down a frictionless incline: comparing solving with kinematics vs. work-energy theorem.

Problem-Solving Strategy: Work-Energy Theorem

1. Draw Free-Body Diagram: Identify all forces acting on the object.
2. Determine Work: Calculate the work done by each force.
   • Be mindful of signs (positive, negative, zero).
3. Calculate Net Work: Sum all individual works: \(W_{net} = \sum W_i\).
4. Apply Theorem: Set \(W_{net} = K_B - K_A = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2\).
   • Solve for the unknown.
5. Check Answer:
   • If \(W_{net} = 0\), speed should be constant.
   • If \(W_{net} > 0\), speed should increase.
   • If \(W_{net} < 0\), speed should decrease.

• This structured approach helps students systematically apply the theorem.
• Reiterate the importance of including all forces for net work.

Example 7.9: Loop-the-Loop

Problem: What minimum height (\(y_1\)) must a toy car start from rest on a frictionless track to complete a loop of radius \(R\)?

Strategy:

1. Forces: Gravity and Normal force.
2. Work: Only gravity does work (normal force is perpendicular to displacement).
3. Apply Work-Energy Theorem from initial height \(y_1\) to the top of the loop (\(y_2 = 2R\)). \(W_{net} = W_{grav} = K_{top} - K_{start}\) \(W_{grav} = -mg(y_2 - y_1) = -mg(2R - y_1)\) \(K_{start} = 0\) (starts from rest) \(K_{top} = \frac{1}{2}mv_{top}^2\) So, \(-mg(2R - y_1) = \frac{1}{2}mv_{top}^2\)
4. Condition for staying on track: At the top, the normal force \(N \ge 0\). Centripetal force at top: \(N + mg = \frac{mv_{top}^2}{R}\). For minimum height, \(N=0\), so \(mg = \frac{mv_{top}^2}{R} \implies v_{top}^2 = gR\).
5. Substitute \(v_{top}^2\) into the Work-Energy equation and solve for \(y_1\).

Figure 7.12: Loop-the-loop track.

Example 7.9: Loop-the-Loop (Solution)

Solution:

From Work-Energy Theorem:

\(-mg(2R - y_1) = \frac{1}{2}mv_{top}^2\)

From condition for staying on track (\(N=0\) at top):

\(v_{top}^2 = gR\)

Substitute \(v_{top}^2\) into the Work-Energy equation:

\(-mg(2R - y_1) = \frac{1}{2}m(gR)\)

Divide by \(mg\):

\(-(2R - y_1) = \frac{1}{2}R\)

\(-2R + y_1 = \frac{1}{2}R\)

\(y_1 = 2R + \frac{1}{2}R\)

\(y_1 = \frac{5}{2}R\)

Significance:

• The minimum starting height is \(2.5R\).
• This result is independent of the car’s mass.
• Demonstrates how the work-energy theorem simplifies problems that would be complex with direct force analysis.

• Walk through the algebraic steps clearly.
• Explain why \(y_1 = 2.5R\) makes physical sense (it’s more than \(2R\) because some kinetic energy is needed to provide centripetal force at the top).
• Emphasize that the beauty of work-energy is bypassing detailed kinematic equations for intermediate points.

7.4 Power

Learning Objectives

By the end of this section, you will be able to:

• Relate the work done during a time interval to the power delivered.
• Find the power expended by a force acting on a moving body.

Defining Power

• Power is the rate at which work is done or energy is transferred.
• Average power (\(P_{avg}\)): \[P_{avg} = \frac{W}{\Delta t}\]
• Instantaneous power (\(P\)): \[P = \frac{dW}{dt}\]

Note

Units of Power:

In SI, power is measured in Watts (W), where \(1\ W = 1\ J/s\).

Another common unit is horsepower (hp): \(1\ hp = 746\ W\).

• Introduce power as a practical extension of work, bringing in the element of time.
• Analogy: two people lift same weight to same height (same work), but one does it faster (more power).
• Explain the difference between average and instantaneous power.
• Connect to real-world examples: car engines, light bulbs, human exertion.

Power and Force-Velocity Relationship

Work as an Integral of Power

• If power varies with time, the work done over an interval is:

  \[W = \int P\ dt\]

Power in terms of Force and Velocity

• For a force \(\vec{F}\) acting on a body with velocity \(\vec{v}\):

  \[P = \vec{F} \cdot \vec{v}\]

• This is the instantaneous power delivered by the force.

  (Derivation: \(P = dW/dt = \vec{F} \cdot d\vec{r}/dt = \vec{F} \cdot \vec{v}\))

Figure 7.14: Pull-up power.

• Explain the \(\vec{F} \cdot \vec{v}\) relationship. This is extremely useful in many physics problems, especially when force and velocity are known.
• Use Figure 7.14 as a simple example: Force is roughly equal to weight, and velocity is the upward speed.

Example 7.11: Pull-Up Power

Problem: An 80-kg army trainee does pull-ups. Takes 0.8 s to raise body from lower position to chin above bar. How much power do muscles supply? (Assume \(\Delta y = 0.6\ m\), and 10% of body mass (arms) not included in moving mass).

Strategy:

1. Calculate work done against gravity: \(W = mg_{effective}\Delta y\).
2. Calculate power: \(P = W/\Delta t\).

Solution:

Effective mass moved: \(m_{effective} = 0.9 \times 80\ kg = 72\ kg\).

Work done: \(W = (72\ kg)(9.8\ m/s^2)(0.60\ m) = 423.36\ J\).

Power: \(P = \frac{423.36\ J}{0.8\ s} = 529.2\ W\).

Significance:

• About 529 W, which is roughly 0.71 horsepower.
• This is a realistic power output for strenuous human activity.
• Highlights the importance of making reasonable approximations in physics problems.

• Discuss the assumptions made (effective mass, vertical displacement).
• Connect to the horsepower unit (\(1\ hp = 746\ W\)) to give a familiar context for the power output.

Example 7.12: Automotive Power Driving Uphill

Problem: How much power must a 1200-kg car engine expend to move up a 15% grade at 90 km/h? Assume 25% of power is dissipated by air resistance and friction.

Strategy:

1. Identify forces: Gravity and Engine thrust.
2. At constant velocity, net work is zero, so power from engine equals power expended against gravity and resistance.
3. Power against gravity (\(P_g\)) is \(F_g \cdot \vec{v} = mgv\sin\theta\).
4. A 15% grade means \(\tan\theta = 0.15\).
5. Engine provides 75% of total required power. So, \(0.75 P_{engine} = P_g + P_{resistance\_useful}\). (Alternatively: \(P_{engine}\) provides \(P_g\) + \(P_{air\_friction}\). The problem says 25% of this power is dissipated. So \(P_{engine} = P_{useful} + P_{dissipated}\). \(P_{dissipated} = 0.25 P_{engine}\). Thus, \(P_{useful} = 0.75 P_{engine}\).)

Solution:

Convert speed: \(v = 90\ km/h = 90 \times \frac{1000\ m}{3600\ s} = 25\ m/s\).

Calculate \(\theta\): \(\theta = \arctan(0.15) \approx 8.53^\circ\).

Power against gravity: \(P_g = mgv\sin\theta = (1200\ kg)(9.8\ m/s^2)(25\ m/s)\sin(8.53^\circ)\).

\(P_g \approx (11760\ N)(25\ m/s)(0.148) \approx 43497\ W\).

This \(P_g\) is the useful power that must come from the engine. If this is 75% of the total engine power:

\(0.75 P_{engine} = P_g\)

\(P_{engine} = \frac{43497\ W}{0.75} \approx 57996\ W \approx 58\ kW\).

\(58\ kW \approx 78\ hp\).

Figure 7.15: Car moving uphill.

Key Takeaways

1. Work is Energy Transfer: Work quantifies the energy transferred to or from an object by a force causing displacement. \(W = \int \vec{F} \cdot d\vec{r}\).
2. Kinetic Energy is Energy of Motion: Kinetic energy (\(K = \frac{1}{2}mv^2\)) is the energy an object possesses due to its motion.
3. Work-Energy Theorem: The net work done on an object equals the change in its kinetic energy: \(W_{net} = \Delta K\). This is a powerful tool for analyzing motion.
4. Conservative vs. Nonconservative Forces:
   • Conservative forces (like gravity, spring force) do work that depends only on the start and end points, not the path. Total work over a closed path is zero.
   • Nonconservative forces (like friction) do work that depends on the path. Total work over a closed path is non-zero.
5. Power is Rate of Work: Power is how fast work is done or energy is transferred (\(P = \frac{dW}{dt}\)). It can also be expressed as \(P = \vec{F} \cdot \vec{v}\).

Key Equations

Equation	Description
\(dW = \vec{F} \cdot d\vec{r}\)	Infinitesimal work done by a force
\(W_{AB} = \int_{path\ AB} \vec{F} \cdot d\vec{r}\)	Total work done by a force over a path
\(W = Fd\cos\theta\)	Work done by a constant force
\(W_{grav} = -mg(y_B - y_A)\)	Work done by gravity (constant force)
\(W_{spring} = -\frac{1}{2}k(x_B^2 - x_A^2)\)	Work done by a spring force (variable force)
\(K = \frac{1}{2}mv^2\)	Kinetic energy of a particle
\(K = \frac{p^2}{2m}\)	Kinetic energy in terms of momentum
\(W_{net} = K_B - K_A = \Delta K\)	Work-Energy Theorem (net work equals change in kinetic energy)
\(P_{avg} = \frac{W}{\Delta t}\)	Average power
\(P = \frac{dW}{dt}\)	Instantaneous power
\(P = \vec{F} \cdot \vec{v}\)	Power in terms of force and velocity

Key Terms

Term	Definition
Work (W)	Energy transferred to or from an object by a force acting over a displacement.
Joule (J)	SI unit of work and energy; \(1\ J = 1\ N \cdot m\).
Kinetic Energy (K)	Energy an object possesses due to its motion.
Work-Energy Theorem	States that the net work done on an object equals the change in its kinetic energy.
Power (P)	The rate at which work is done or energy is transferred.
Watt (W)	SI unit of power; \(1\ W = 1\ J/s\).
Horsepower (hp)	Common unit of power, \(1\ hp = 746\ W\).
Conservative Force	A force for which the work done is independent of the path taken (e.g., gravity, spring force).
Nonconservative Force	A force for which the work done depends on the path taken (e.g., friction, air resistance).
Spring Constant (k)	A measure of the stiffness of a spring, relating force to displacement (Hooke’s Law).