Chapter 6Comprehensive Study Guide to Work, Energy, and Power

Work and Mechanical Energy Transfer

• Definition of Work: In physics, work is defined as the transfer of energy from one object to another by mechanical means. A force or a combination of forces must act on an object over a displacement for energy to be added to or removed from that object.

• Calculation of Work Done by a Constant Force: The work $W$ performed on an object by a constant force is the product of the magnitude of the displacement $d$ and the parallel component of the force in the direction of motion $F_{||}$.

• Mathematical Formulations:

  • $W = F_{||} d$

  • $F_{||} = F \cos(\theta)$

  • $W = (F \cos(\theta)) d = F d \cos(\theta)$

  • In these equations, $F$ is the magnitude of the applied force, $d$ is the magnitude of the displacement, and $\theta$ represents the smaller angle between the force vector and the displacement vector when placed tail to tail.

• Scalar Nature and Sign Conventions of Work:

  • Work is a scalar quantity; it possesses magnitude but no physical direction in space, even though it is calculated from two vector quantities ($\vec{F}$ and $\vec{d}$).

  • Positive Work ($\cos(\theta) > 0$, where $0^\circ \le \theta < 90^\circ$): Indicates that mechanical energy (kinetic or potential energy) is being transferred into the object or system.

  • Negative Work ($\cos(\theta) < 0$, where $90^\circ < \theta \le 180^\circ$): Indicates that mechanical energy is being transferred out of or removed from the object or system.

  • Zero Work ($\cos(\theta) = 0$, where $\theta = 90^\circ$): Occurs when the applied force is perpendicular to the displacement. No mechanical energy is transferred by that force.

  • Zero Work from Lack of Displacement ($d = 0$): Even if a large force is exerted (such as pushing against an immovable wall or holding a heavy object stationary), no mechanical work is performed because the displacement is zero.

• Perpendicular Forces in Motion:

  • When a person carries a package of mass $m$ walking horizontally at constant velocity, the upward supporting force $F_P$ equals $mg$, but the displacement $\vec{d}$ is horizontal. Because $\theta = 90^\circ$, $\cos(90^\circ) = 0$, and the work done by the upward supporting force is exactly $0\,\text{J}$.

  • Similarly, the gravitational force exerted by the Earth on the Moon acts inward toward the center of the orbit (centripetally), perpendicular to the Moon's instantaneous tangential orbital velocity vector. Therefore, the gravitational force of the Earth does no work on the Moon ($W = 0\,\text{J}$).

Kinetic Energy and the Work-Energy Principle

• Kinetic Energy ($KE$): The energy possessed by an object by virtue of its motion.

• Kinetic Energy Formula:

  • $KE = \frac{1}{2} m v^2 = \frac{m v^2}{2}$

  • where $m$ is the mass of the object and $v$ is its instantaneous speed.

• Change in Kinetic Energy ($\Delta KE$): For an object with constant mass $m$ undergoing a change in speed from initial velocity $v_i$ to final velocity $v_f$:

  • $\Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$

• The Work-Energy Principle: The net work done on an object by the vector sum of all forces acting on it ($W_{net}$) equals the net change in the object's kinetic energy:

  • $W_{net} = \Delta KE$

• Net Work vs. Individual Work Distinction:

  • The equation $W = \Delta KE$ does not hold true for arbitrary individual forces; it strictly requires the net work $W_{net}$.

  • If $W_{net} > 0$, the kinetic energy increases ($KE_f > KE_i$).

  • If $W_{net} < 0$, the kinetic energy decreases ($KE_f < KE_i$).

  • If $W_{net} = 0$, the kinetic energy remains unchanged ($KE_f = KE_i$).

• Nonlinear Speed-to-Energy Relationship and Stopping Distances:

  • Because kinetic energy is proportional to the square of the speed ($KE \propto v^2$), doubling the speed quadruples ($2^2 = 4$) the kinetic energy.

  • Under a constant braking friction force $F$, the work done to bring a vehicle to a complete stop ($v_2 = 0$) is:

  • $W_{net} = -F d = 0 - KE_1 = -\frac{1}{2} m v_1^2$

  • $d = \frac{m v_1^2}{2 F}$

  • Consequently, the stopping distance is proportional to the square of the initial speed ($d \propto v^2$).

Conservative Forces, Nonconservative Forces, and Potential Energy

• Classification of Forces: All forces in mechanics are classified into two broad categories: conservative forces and nonconservative forces.

• Conservative Forces:

  • Definition: A force is conservative if the work done by the force on an object moving between two points depends only on the initial and final positions, and is completely independent of the path taken.

  • Closed Path Characteristic: The work done by a conservative force around any closed path is zero ($\oint \vec{F} \cdot d\vec{r} = 0$).

  • Potential Energy Definition: Potential energy ($PE$ or $U$) can only be defined for conservative forces.

  • Examples of Conservative Forces:

  • Gravitational force

  • Elastic (spring) force

  • Electric force

• Nonconservative Forces:

  • Definition: A force is nonconservative if the work done by the force depends explicitly on the path taken between the starting and ending points.

  • Path Dependence Illustration: When pushing a crate across a floor between points 1 and 2, a curved or longer path requires more total work against friction than a direct straight path, because friction continually opposes the instantaneous direction of motion.

  • Examples of Nonconservative Forces:

  • Friction

  • Air resistance

  • Tension in a cord

  • Motor or rocket propulsion

  • Push or pull exerted by a person

Gravitational and Elastic Potential Energy

• Potential Energy ($PE$ or $U$): The energy stored in a system by virtue of the relative position, state, or configuration of its parts in interaction with their surroundings.

  • Familiar examples include a wound-up spring, a stretched elastic band, or an object elevated to a height above the ground.

  • Potential energy is fundamentally a property of an entire interacting system, not of an isolated object alone (e.g., gravitational potential energy belongs to the object-Earth system).

• Gravitational Potential Energy ($PE_g$):

  • Defined as the work done by an external force against gravity to change the vertical elevation of an object of mass $m$ without acceleration:

  • $W_{ext} = F_{ext} d \cos(0^\circ) = m g h = m g (y_2 - y_1)$

  • The resulting change in gravitational potential energy is:

  • $\Delta PE_g = m g \Delta y = m g \Delta h$

  • Arbitrary Nature of Reference Level ($PE_g = 0$):

  • When writing $PE_g = mgh$, the zero reference elevation ($h = 0$) can be chosen arbitrarily at any convenient point in the coordinate system.

  • Only changes in potential energy ($\Delta PE_g$) can be physically measured or observed. As long as the chosen reference plane remains consistent throughout the entire calculation, the final kinematic and dynamic results are identical.

• Elastic (Spring) Potential Energy ($PE_s$):

  • Hooke's Law: The magnitude of the external force $F_P$ required to stretch or compress an ideal spring by a displacement $x$ from its equilibrium position ($x = 0$) is linearly proportional to $x$:

  • $F_P = k x$

  • where $k$ is the spring constant (a measure of spring stiffness, measured in $\text{N/m}$), which must be determined empirically for each individual spring.

  • The restoring force exerted by the spring ($F_s$) acts in the opposite direction: $F_s = -k x$.

• Work Done to Compress or Stretch a Spring:

  • Because the force varies linearly from $0$ at equilibrium to $k x_f$ at final displacement $x_f$, the average force is:

    • $\bar{F} = \frac{1}{2} k x_f$

  • The work performed equals the triangular area beneath the force versus displacement plot:

    • $W = \bar{F} x_f = \left(\frac{1}{2} k x_f\right) x_f = \frac{1}{2} k x_f^2$

  • Therefore, the potential energy stored in an elastic spring ($PE_s$) stretched or compressed by a distance $x$ from equilibrium is:

    • $PE_s = \frac{1}{2} k x^2$

Conservation of Mechanical Energy

• Total Mechanical Energy ($E$): Defined as the sum of all kinetic and potential energies within a system:

  • $E = KE + PE$

• Derivation of the General Energy Equation:

  • Partition total work into work done by conservative forces ($W_c$) and work done by nonconservative forces ($W_{nc}$):

  • $W_{net} = W_c + W_{nc}$

  • From the Work-Energy Principle:

  • $W_c + W_{nc} = \Delta KE$

  • $W_{nc} = \Delta KE - W_c$

  • By definition, the work done by conservative forces equals the negative change in potential energy of the system ($W_c = -\Delta PE$):

  • $W_{nc} = \Delta KE - (-\Delta PE)$

  • $W_{nc} = \Delta KE + \Delta PE$

  • Expanding the change terms into initial ($i$) and final ($f$) states:

  • $W_{nc} = (KE_f - KE_i) + (PE_f - PE_i)$

  • $KE_i + PE_i + W_{nc} = KE_f + PE_f$

• Conservation Law for Isolated/Conservative Systems:

  • When no nonconservative forces act on the system, or when their effects are negligible ($W_{nc} = 0$):

  • $0 = \Delta KE + \Delta PE$

  • $\Delta E = 0$

  • $KE_i + PE_i = KE_f + PE_f$

  • Verbal Statement: The total mechanical energy of an isolated system remains constant throughout any process when only conservative forces do work within the system.

Conservation of Energy with Nonconservative Forces and Friction

• Inclusion of Nonconservative Work:

  • When nonconservative forces (such as applied pushes, motor output, tension, or friction) do work, mechanical energy is added to or dissipated from the mechanical system:

  • $KE_i + PE_i + W_{nc} = KE_f + PE_f$

• Work Done by Kinetic Friction:

  • Kinetic friction always acts in the direction directly opposite to the instantaneous relative motion, making the angle between friction force and displacement vector $\theta = 180^\circ$.

  • Because $\cos(180^\circ) = -1$, the work done by kinetic friction is always negative:

  • $W_{fr} = F_{fr} d \cos(180^\circ) = -f_k d$

  • where $f_k = \mu_k F_N$ ($\mu_k$ is the coefficient of kinetic friction and $F_N$ is the normal force).

• Mechanical Energy Equation Explicitly Accounting for Friction:

  • Substituting $W_{nc} = -f_k d$ yields:

  • $KE_i + PE_i - f_k d = KE_f + PE_f$

  • The energy term $f_k d$ represents the mechanical energy converted into non-recoverable thermal energy (heat).

Power: Definition, Units, and Calculations

• Definition of Power: Power is the time rate at which work is performed, or the time rate at which energy is transformed or transferred.

  • Energy cannot be created or destroyed; thus, power describes the rate at which energy leaves a defined system or is converted from one form to another.

• Average Power Formula:

  • $P_{ave} = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$

  • Or, using simplified notation where $t$ represents elapsed time:

  • $P_{ave} = \frac{W}{t} = \frac{E}{t}$

• Alternative Power Formula in Terms of Velocity:

  • By substituting constant force work into the power definition:

  • $P = \frac{W}{t} = \frac{F d \cos(\theta)}{t} = F \left(\frac{d}{t}\right) \cos(\theta)$

  • $P = F v_{ave} \cos(\theta)$

  • Choice between $P = \frac{W}{t}$ and $P = F v_{ave} \cos(\theta)$ depends on whether total work/energy or force and velocity are known.

• SI Unit of Power:

  • The SI unit of power is the Watt ($\text{W}$):

  • $1\,\text{Watt} = 1\,\text{W} = 1\,\text{J/s} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^3$

• Disambiguation of the Symbol "W":

  • $W = m g$: Here, $W$ represents Weight, which is a downward gravitational force measured in Newtons ($\text{N}$).

  • $W = F d \cos(\theta)$: Here, $W$ represents Work, which is an energy transfer measured in Joules ($\text{J}$).

  • $3\,\text{W}$: Here, $\text{W}$ is the unit symbol for Watts, representing a power of $3\,\text{J/s}$.

• Energy Consumed by Electrical Appliances:

  • From $P_{ave} = \frac{E}{t}$, the energy consumed over operating time $t$ is:

  • $E_{used} = P_{ave} t$

• Commercial Unit of Energy — The Kilowatt-Hour ($\text{kW-hr}$):

  • While the Joule ($\text{J}$) is the standard SI unit of energy, domestic electrical utility billing uses the kilowatt-hour ($\text{kW-hr}$) to avoid unwieldy numbers (such as tens of billions of Joules).

  • Power vs. Energy Units: A kilowatt ($\text{kW}$) is a unit of power ($1\,\text{kW} = 1000\,\text{W}$). A kilowatt-hour ($\text{kW-hr}$) is a unit of total energy.

  • Conversion Between Joules and Kilowatt-Hours:

  • $E = (1\,\text{kW})(1\,\text{hr}) = (1000\,\text{W})(3600\,\text{s}) = 3,600,000\,\text{J} = 3.60 \times 10^6\,\text{J}$

  • $1\,\text{kW-hr} = 3.60 \times 10^6\,\text{J}$

Questions & Discussion

• Question: What is the nature of an electric utility company?

  • Prompt: The local electric company is: (A) A power company, (B) An energy company, (C) Either.

  • Resolution: Either designation is commonly used colloquially, but scientifically it delivers energy measured in kilowatt-hours over time, which relates to power output.

• Question: Which definition correctly describes power?

  • Prompt: Power is: 1. The rate that work is done, 2. The rate energy is used, 3. Both of the above, 4. Neither of the above.

  • Resolution: Option 3 (Both of the above). Power measures both the rate of work done and the rate of energy transformation.

• Worked Example: Average Power Output of a Runner on Stairs

  • Problem: A $128\,\text{lb}$ ($570\,\text{N}$) woman runs up a flight of stairs having a vertical rise of $14\,\text{ft}$ ($4.3\,\text{m}$) in $3.5\,\text{s}$. What average power must she supply?

  • Solution Steps:

  • The work she performs is converted directly to gravitational potential energy:

    • $W = \Delta PE_g = m g h = (570\,\text{N})(4.3\,\text{m}) = 2451\,\text{J}$

  • Compute average power:

    • $P_{ave} = \frac{W}{t} = \frac{2451\,\text{J}}{3.5\,\text{s}} = 700.28\,\text{W} \approx 700\,\text{W}$

  • Answer: $700\,\text{W}$.

• Worked Example: Joules Consumed by a Lightbulb

  • Problem: How many Joules are consumed by a $100\,\text{W}$ bulb operated continuously for $24\,\text{hours}$?

  • Solution Steps:

  • Convert time to seconds: $t = (24\,\text{hr}) \times (3600\,\text{s/hr}) = 86,400\,\text{s}$

  • Calculate energy in Joules:

    • $E = P t = (100\,\text{W})(86,400\,\text{s}) = 8,640,000\,\text{J} = 8.64 \times 10^6\,\text{J}$

  • Answer: $8,640,000\,\text{J}$ (or $8.64\,\text{MJ}$, equal to $2.4\,\text{kW-hr}$ calculated as $E = \left(\frac{100}{1000}\,\text{kW}\right)(24\,\text{hr}) = 2.4\,\text{kW-hr}$).

• Worked Example: Hill Climbing Power of a Cyclist

  • Problem: How fast must a cyclist climb a hill with a $6^\circ$ incline to maintain a power output of $0.25\,\text{hp}$? Neglect work done by friction and assume the combined mass of the bicycle and cyclist is $68\,\text{kg}$.

  • Solution Steps:

  • Convert horsepower to Watts ($1\,\text{hp} = 746\,\text{W}$):

    • $P = 0.25\,\text{hp} \times 746\,\text{W/hp} = 186.5\,\text{W}$

  • The component of gravitational force acting parallel down the incline is:

    • $F_{||} = m g \sin(\theta) = (68\,\text{kg})(9.8\,\text{m/s}^2) \sin(6^\circ) = (666.4\,\text{N})(0.1045) \approx 69.66\,\text{N}$

  • At constant speed, applied force equals the gravitational resistance force ($F = F_{||}$):

    • $P = F v \implies v = \frac{P}{F} = \frac{186.5\,\text{W}}{69.66\,\text{N}} \approx 2.68\,\text{m/s}$

  • Answer: $2.68\,\text{m/s}$.

• Question: Kinetic Energy Comparison for Accelerating Automobile

  • Prompt: Before the brakes are applied, how does the kinetic energy of a car traveling at $120\,\text{km/h}$ compare to its kinetic energy at $60\,\text{km/h}$?

  • Resolution: The speed is doubled ($\frac{120}{60} = 2$). Because $KE \propto v^2$, the kinetic energy is $2^2 = 4$ times as much.

• Question: Vehicle Stopping Distance with Constant Braking Force

  • Prompt: Assuming the braking force is constant, if a car traveling at $60\,\text{km/h}$ stops in $20\,\text{m}$, what is the stopping distance when the speed is doubled to $120\,\text{km/h}$?

  • Resolution: Because stopping distance $d \propto v^2$, doubling the speed increases the stopping distance by a factor of $4$:

  • $d = 4 \times 20\,\text{m} = 80\,\text{m}$

• Question: Energy Transformation in Pole Vaulting

  • Prompt: At the instant a pole vaulter clears the crossbar, he is almost stationary. What happened to all the kinetic energy he possessed before planting his pole?

  • Resolution: It is stored as gravitational potential energy in the system of the man and the Earth (as well as momentarily in the elastic deformation of the pole during the rise).

• Question: True or False Conceptual Checks on Work, Force, and Energy

  • Prompt 1: If a net force does work on an object, the $KE$ of the object will increase.

  • Answer: Sometimes true. If the net work is positive, $KE$ increases; if the net work is negative, $KE$ decreases.

  • Prompt 2: Forces can be divided into two types, conservative and non-conservative.

  • Answer: True.

  • Prompt 3: Potential energy can be defined only for a conservative force.

  • Answer: True.

• Worked Example: Vertical Spring Launcher

  • Problem: A $0.2\,\text{kg}$ mass is pushed down on a vertical spring ($k = 400\,\text{N/m}$). The spring is compressed by $0.05\,\text{m}$. Neglecting the mass of the spring and all nonconservative forces, calculate: (a) how high the mass will rise above the compressed position when released, and (b) how fast the mass is moving at the instant it leaves the spring.

  • Solution Steps for (a):

  • Initial state ($i$): spring compressed by $x = 0.05\,\text{m}$, mass at rest ($KE_i = 0$), set initial height $y_i = 0$.

    • $E_i = \frac{1}{2} k x^2 = \frac{1}{2}(400\,\text{N/m})(0.05\,\text{m})^2 = 0.5\,\text{J}$

  • Peak state ($f$): mass at highest elevation $h$, instantaneously at rest ($KE_f = 0$), spring uncompressed ($PE_{s,f} = 0$):

    • $E_f = m g h = (0.2\,\text{kg})(9.8\,\text{m/s}^2) h = 1.96 h$

  • Equating initial and final mechanical energy:

    • $0.5 = 1.96 h \implies h = \frac{0.5}{1.96} \approx 0.255\,\text{m}$

  • Solution Steps for (b):

  • Instant of release from spring: spring returns to equilibrium ($x = 0$, so $PE_s = 0$), vertical elevation gained is $y = 0.05\,\text{m}$:

    • $E_i = E$

    • $0.5\,\text{J} = m g y + \frac{1}{2} m v^2$

    • $0.5 = (0.2\,\text{kg})(9.8\,\text{m/s}^2)(0.05\,\text{m}) + \frac{1}{2}(0.2\,\text{kg}) v^2$

    • $0.5 = 0.098 + 0.1 v^2$

    • $0.1 v^2 = 0.402 \implies v^2 = 4.02 \implies v = \sqrt{4.02} \approx 2.01\,\text{m/s}$

  • Answers: (a) $h = 0.255\,\text{m}$, (b) $v = 2.01\,\text{m/s}$.

• Worked Example: The Comprehensive Pendulum Problem

  • Problem: A pendulum with a mass of $1.60\,\text{kg}$ hangs at rest $1.0\,\text{m}$ above the floor. It is displaced to a height of $1.75\,\text{m}$ above the floor and released from rest.

  1. Possible sources of friction: Air friction (drag) and mechanical friction at the pivot support at the top of the string.

  2. Maximum gravitational potential energy ($PE_g$): Taking the lowest point ($1.0\,\text{m}$ above the floor) as $PE_g = 0$, the change in height is $\Delta h = 1.75\,\text{m} - 1.0\,\text{m} = 0.75\,\text{m}$.

     • $PE_{g,max} = m g \Delta h = (1.60\,\text{kg})(9.8\,\text{m/s}^2)(0.75\,\text{m}) = 11.76\,\text{J} \approx 11.8\,\text{J}$

  3. Location and value of maximum kinetic energy: Kinetic energy is maximized at the lowest point of the swing where all potential energy has converted into kinetic energy:

     • $KE_{max} = 11.76\,\text{J} \approx 11.8\,\text{J}$

  4. Height reached upon completion of one full swing: It returns to the exact same release height ($1.75\,\text{m}$ above floor) due to conservation of mechanical energy.

  5. Maximum speed: Occurs at the bottom of the swing:

     • $\frac{1}{2} m v_{max}^2 = 11.76\,\text{J} \implies v_{max} = \sqrt{\frac{2(11.76\,\text{J})}{1.60\,\text{kg}}} = \sqrt{14.7} \approx 3.83\,\text{m/s}$

  6. Speed after dropping $0.50\,\text{m}$ below the release point:

     • $\Delta PE_g = m g \Delta y = (1.60)(9.8)(0.50) = 7.84\,\text{J}$

     • $v = \sqrt{2 g \Delta y} = \sqrt{2(9.8\,\text{m/s}^2)(0.50\,\text{m})} = \sqrt{9.8} \approx 3.13\,\text{m/s}$

  7. Behavior when given an initial velocity of $0.40\,\text{m/s}$ at $1.75\,\text{m}$ elevation:

     • $E_i = m g h_i + \frac{1}{2} m v_i^2 = 11.76\,\text{J} + \frac{1}{2}(1.60\,\text{kg})(0.40\,\text{m/s})^2 = 11.76 + 0.128 = 11.888\,\text{J}$

     • Final height above lowest equilibrium point:

       • $h = \frac{E_i}{m g} = \frac{11.888\,\text{J}}{(1.60\,\text{kg})(9.8\,\text{m/s}^2)} = \frac{11.888}{15.68} \approx 0.758\,\text{m}$

     • Height above lowest point upon return: $0.758\,\text{m}$ by conservation of mechanical energy.

• Question: Vertical Spring Dart Gun Compression Scaling

  • Prompt: A spring-loaded dart gun fired vertically shoots a dart to a maximum height of $24\,\text{m}$. If fired a second time with the spring compressed only half as far ($x' = \frac{x}{2}$), how high does the dart rise?

  • Resolution: Potential energy stored in a spring is proportional to the square of compression ($PE_s \propto x^2$). Halving the compression reduces the available initial energy by a factor of $4$ ($\left(\frac{1}{2}\right)^2 = \frac{1}{4}$). Since maximum vertical height is directly proportional to initial energy ($mgh = PE_s$), the new height is:

  • $h' = \frac{24\,\text{m}}{4} = 6\,\text{m}$

• Question: Trajectory Independence of Impact Speed

  • Prompt: Two identical balls are thrown off a building of height $h$ with the same initial speed $v_0$, one thrown vertically upward and one thrown at a $45^\circ$ angle. If air resistance is neglected, which hits with greater speed?

  • Resolution: Both hit the ground with identical speed. By conservation of mechanical energy:

  • $m g h + \frac{1}{2} m v_0^2 = 0 + \frac{1}{2} m v_f^2 \implies v_f = \sqrt{v_0^2 + 2 g h}$

  • The impact speed depends solely on initial speed and initial elevation, not on trajectory launch angle (although they strike the ground at different times).

• Worked Example: Dropped Baseball with Air Friction

  • Problem: A $145\,\text{g}$ ($0.145\,\text{kg}$) baseball is dropped from rest from a height of $13.0\,\text{m}$.

  • (a) Calculate the speed when it reaches the ground ignoring friction.

  • (b) Calculate the average air friction force acting on the ball if the actual final speed is measured to be $15.0\,\text{m/s}$.

  • Solution Steps for (a):

  • $m g h = \frac{1}{2} m v^2 \implies v = \sqrt{2 g h} = \sqrt{2(9.8\,\text{m/s}^2)(13.0\,\text{m})} = \sqrt{254.8} \approx 15.96\,\text{m/s} \approx 16.0\,\text{m/s}$

  • Solution Steps for (b):

  • Apply mechanical energy conservation with friction work ($W_{nc} = -f_{air} d$):

    • $m g h - f_{air} d = \frac{1}{2} m v_f^2$

    • Initial energy: $E_i = (0.145\,\text{kg})(9.8\,\text{m/s}^2)(13.0\,\text{m}) = 18.473\,\text{J}$

    • Final kinetic energy: $KE_f = \frac{1}{2}(0.145\,\text{kg})(15.0\,\text{m/s})^2 = 16.3125\,\text{J}$

    • Energy lost to friction: $f_{air} d = 18.473\,\text{J} - 16.3125\,\text{J} = 2.1605\,\text{J}$

    • Average air friction force: $f_{air} = \frac{2.1605\,\text{J}}{13.0\,\text{m}} \approx 0.166\,\text{N}$

  • Answers: (a) $16.0\,\text{m/s}$, (b) $0.166\,\text{N}$.

• Worked Example: Block Sliding Up an Incline with Friction

  • Problem: A $4.0\,\text{kg}$ block starts at the bottom of a $30^\circ$ incline with an initial kinetic energy of $128\,\text{J}$. Calculate how far ($d$) it will slide up the plane if the coefficient of kinetic friction is $\mu_k = 0.30$.

  • Solution Steps:

  • Gravitational normal force on incline: $F_N = m g \cos(30^\circ)$

  • Kinetic friction force: $f_k = \mu_k F_N = \mu_k m g \cos(30^\circ)$

  • Vertical height reached along displacement $d$: $h = d \sin(30^\circ)$

  • Energy balance equation ($KE_f = 0$ at stopping point):

    • $KE_i - f_k d = m g h$

    • $KE_i = m g d \sin(30^\circ) + \mu_k m g d \cos(30^\circ)$

    • $KE_i = m g d [\sin(30^\circ) + \mu_k \cos(30^\circ)]$

  • Plug in values:

    • $m g = (4.0\,\text{kg})(9.8\,\text{m/s}^2) = 39.2\,\text{N}$

    • $\sin(30^\circ) = 0.5000$

    • $\cos(30^\circ) \approx 0.8660$

    • $\sin(30^\circ) + \mu_k \cos(30^\circ) = 0.5000 + (0.30)(0.8660) = 0.5000 + 0.2598 = 0.7598$

    • $128\,\text{J} = (39.2\,\text{N}) d (0.7598) = 29.784 d$

    • $d = \frac{128}{29.784} \approx 4.298\,\text{m} \approx 4.3\,\text{m}$

  • Answer: $4.3\,\text{m}$.