Introductory Physics: Energy, Work, and Motion

Fundamental Concepts of Mechanical Energy

• Two Main Types of Energy:

  • Kinetic Energy ($KE$ or $K$): Energy associated with an object in motion. An object at rest ($v = 0$) has zero kinetic energy.
  • Formula: $KE = \frac{1}{2} m v^2$
  • $m$ = mass of the object in kilograms ($kg$)
  • $v$ = speed or velocity of the object in meters per second ($m/s$)
  • Potential Energy ($PE$ or $U$): Stored energy due to an object's position relative to a reference height.
  • Formula: $PE = m g h$
  • $m$ = mass of the object in kilograms ($kg$)
  • $g$ = acceleration due to gravity, taken standardly as $9.80View$ or $9.80\,m/s^2$
  • $h$ = vertical height relative to a surface (floor, table, chair, or ground)

• Conservation of Mechanical Energy:

  • Total mechanical energy $E$ is the sum of kinetic and potential energy:     $E = KE + PE$
  • In the absence of non-conservative forces (such as friction or air resistance), total mechanical energy $E$ remains constant throughout the motion.

Conservation of Energy Demonstrations and Scenarios

• Flagpole Diving Scenario (Demonstrated by Professor Hewitt):

  • A diver stands at rest at the top of a high flagpole before diving into a bucket of water on the ground.
  • Top of the Flagpole (At Rest, $v = 0$):
  • Potential Energy: $PE = 10{,}000\,\text{units}$
  • Kinetic Energy: $KE = 0\,\text{units}$
  • Total Mechanical Energy: $E = 10{,}000\,\text{units}$
  • Halfway Down the Fall:
  • Height is halved, so potential energy is halved: $PE = 5{,}000\,\text{units}$
  • Kinetic Energy: $KE = 5{,}000\,\text{units}$
  • Total Mechanical Energy: $E = 10{,}000\,\text{units}$
  • Three-Quarters of the Way Down ($\frac{1}{4}$ Height Remaining):
  • Height is one-quarter of initial, so potential energy is: $PE = 2{,}500\,\text{units}$
  • Kinetic Energy: $KE = 7{,}500\,\text{units}$
  • Total Mechanical Energy: $E = 10{,}000\,\text{units}$
  • Bottom of the Fall (At Bucket Level, $h = 0$):
  • Potential Energy: $PE = 0\,\text{units}$
  • Kinetic Energy just before impact: $KE = 10{,}000\,\text{units}$
  • Impact with Water ("Splat"):
  • Upon coming to rest in the bucket, kinetic energy goes to zero ($KE = 0\,\text{units}$).
  • The $10{,}000\,\text{units}$ of kinetic energy is completely transformed into thermal energy (heat of the bucket and remains), which dissipates into the environment.

• Vertical Marker Toss Scenario:

  • A marker is thrown straight up into the air with an initial energy of $3\,J$.
  • At Release Point (Hand Level Reference):
  • Kinetic Energy: $KE = 3\,J$
  • Potential Energy: $PE = 0\,J$
  • At Highest Point of Trajectory:
  • The marker momentarily comes to rest ($v = 0$).
  • Kinetic Energy: $KE = 0\,J$
  • Potential Energy: $PE = 3\,J$
  • When Falling Back to Initial Hand Level:
  • Kinetic Energy: $KE = 3\,J$
  • Potential Energy: $PE = 0\,J$

Step-by-Step Energy Calculations for Falling Objects

• Problem Framework:

  • Acceleration due to gravity value: $g = 9.8\,m/s^2$
  • Air resistance is neglected, making total energy conservative.

• Calculations across Heights ($h_1$ to $h_4$):

  • At Initial Height $h_1$ (Moment of Release):
  • Object is released from rest, so $KE_1 = 0\,J$
  • Calculated Potential Energy: $PE_1 = 9.8\,J$
  • Total Energy: $E = KE_1 + PE_1 = 0\,J + 9.8\,J = 9.8\,J$
  • At Height $h_2$:
  • Given Potential Energy: $PE_2 = 6.86\,J$
  • Without knowing speed $v$, kinetic energy is calculated via conservation of energy:       $E = KE_2 + PE_2$$9.8\,J = KE_2 + 6.86\,J$$KE_2 = 9.8\,J - 6.86\,J = 2.94\,J$
  • At Height $h_3$:
  • Potential Energy decreases further, giving a kinetic energy of: $KE_3 = 2.940\,J$
  • At Height $h_4$ (Ground Impact, $h = 0$):
  • Potential Energy: $PE_4 = 0\,J$
  • Kinetic Energy: $KE_4 = 9.8\,J$

• Energy Transformation Trends:

  • As the object falls, energy transfers continuously from potential to kinetic energy.
  • Speed $v$ increases during free fall, causing $KE$ to increase as $PE$ decreases.
  • The absolute maximum energy ceiling remains $9.8\,J$ at every point during motion.

Work, Power, and Physical Units

• Standard Physical Units:

  • Force: Newton ($N$), where $1\,N = 1\,kg \cdot m/s^2$
  • Energy and Work: Joule ($J$), where $1\,J = 1\,N \cdot m = 1\,kg \cdot m^2/s^2$
  • Power: Watt ($W$), where $1\,W = 1\,J/s$

• Work ($W$):

  • Mechanical work represents energy transfer by applying a force over a displacement:     $W = F \times d$
  • Condition for Work: The object must move in the direction of the force. Applying force to a stationary object (e.g., pushing a fixed desk) results in zero mechanical work ($W = 0\,J$).

• Power ($P$):

  • Power is defined as the rate of energy transfer or work done per unit time:     $P = \frac{W}{t} = \frac{F \times d}{t}$

• Sample Calculation 1 (Power Evaluation):

  • Given Data:
  • Force $F = 150\,N$
  • Distance $d = 10\,m$
  • Time $t = 20\,s$
  • Solution:$\text{Work } W = F \times d = 150\,N \times 10\,m = 1500\,J$$\text{Power } P = \frac{W}{t} = \frac{1500\,J}{20\,s} = 75\,W$

Newton's Second Law and Kinematics Problems

• Net Force ($F_{\text{net}}$):

  • The net force on an object is the vector sum of all applied forces acting on it (not merely its weight).

• Problem 1: Pushing a Box against Friction

  • Given Data:
  • Applied pushing force: $F_{\text{push}} = 40\,N$
  • Friction force: $F_{\text{friction}} = 24\,N$
  • Net Force Calculation:$F_{\text{net}} = F_{\text{push}} - F_{\text{friction}} = 40\,N - 24\,N = 16\,N$
  • Acceleration Calculation:$a = \frac{F_{\text{net}}}{m}$     Yielding an acceleration of $a = 4\,m/s^2$.
  • Time to Reach Target Speed:
  • Target speed $v_f = 8\,m/s$
  • Initial speed $v_i = 0\,m/s$ (starts from rest)
  • Acceleration formula: $a = \frac{v_f - v_i}{t_f - t_i}$
  • Calculation:       $4\,m/s^2 = \frac{8\,m/s - 0\,m/s}{t - 0\,s}$$t = \frac{8\,m/s}{4\,m/s^2} = 2\,s$

Impulse, Momentum, and Motion Graphing

• Problem 2: Accelerating Race Car

  • Given Data:
  • Mass of car $m = 800\,kg$
  • Initial speed $v_i = 0\,m/s$ (starts from rest)
  • Final speed $v_f = 2.8\,m/s$
  • Time interval $\Delta t = 0.03\,s$
  • Part A: Final Momentum ($p_f$)
  • Formula: $p = m \times v$
  • Initial momentum: $p_i = 800\,kg \times 0\,m/s = 0\,kg \cdot m/s$
  • Final momentum:       $p_f = 800\,kg \times 2.8\,m/s = 2240\,kg \cdot m/s$
  • Part B: Engine Force ($F$) via Impulse-Momentum Theorem
  • Impulse formula:       $\text{Impulse} = F \times \Delta t = \Delta p = p_f - p_i$
  • Calculation:       $F \times 0.03\,s = 2240\,kg \cdot m/s - 0\,kg \cdot m/s$$F = \frac{2240\,kg \cdot m/s}{0.03\,s} = 74{,}666.67\,N$

• Graphical Representation of Motion:

  • Velocity vs. Time for Constant Acceleration:
  • A straight, tilted/slanted line at an angle.
  • Slope is constant, where $\text{slope} = \frac{\text{rise}}{\text{run}} = a$
  • Velocity vs. Time for Constant Velocity:
  • A flat, horizontal straight line with zero slope.

Questions & Discussion

• Gravity Standard Values:

  • Question: Should gravity $g$ be kept strictly at $9.8\,m/s^2$ or changed?
  • Answer: Use $9.8\,m/s^2$ or $9.80\,m/s^2$ as listed in standard tables. Values like $9.81\,m/s^2$ resulting from different calculator precision are acceptable as long as figures remain close.

• Calculating Kinetic Energy without Velocity:

  • Question: How do you find $KE$ at an intermediate height if velocity is unknown?
  • Answer: Subtract potential energy from total mechanical energy ($KE = E - PE$). Since total mechanical energy is conserved ($9.8\,J$), velocity is not required.

• Handling Frictional Forces in Acceleration Problems:

  • Question: Why subtract friction force from pushing force?
  • Answer: Friction acts in opposition to applied motion, causing deceleration. Subtracting frictional force ($24\,N$) from pushing force ($40\,N$) gives net force ($16\,N$).

• Setting Initial Conditions in Kinematics:

  • Question: What are initial values for time and speed when an object starts moving?
  • Answer: An object starting from rest has initial speed $v_i = 0\,m/s$ and initial time $t_i = 0\,s$.

• Assessment and Laboratory Protocol:

  • Practice unit conversions (grams to kilograms, centimeters to meters) during lab sessions.
  • Submitted problem papers must include complete written structure: Data, Strategy, and Solution steps, rather than unformatted numeric values.