Linear Kinetics, Work, Power, and Energy

Foundations of Linear Kinetics and Newton's Laws of Motion

• Kinetics is the branch of dynamics concerned with forces and their effects on motion.
• Questions addressed by linear kinetics include:
  • What can people do to improve traction when walking on icy streets?
  • Why do some balls bounce higher on one surface than on another?
  • How can football linemen push larger opponents backward?
• Sir Isaac Newton (1642–1727) formulated the fundamental laws governing linear kinetics.
• Newton's First Law of Motion (Law of Inertia):
  • Verbatim Definition: Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.
  • Functional Explanation: An object will not move if it was not moving to begin with, or it will continue moving at a constant speed in a straight line if it was moving to begin with, unless an external force acts on the object.
  • Earth Context: Gravitational force, friction, and air resistance act as external forces on Earth, modifying constant velocity or rest states.
  • Horizontal Motion of a Projectile:
  • Newton's First Law applies to the horizontal motion of a projectile because gravity acts strictly in the vertical direction, assuming air resistance is negligible.
  • A projectile is defined as an object that has no external forces acting on it other than gravity.
  • Horizontal velocity of a projectile is constant and horizontal motion occurs in a straight line (it will not slow down or speed up horizontally once released).
  • If air resistance is significant, the object cannot strictly be classified as a pure projectile.
  • The human body can act as a projectile during movements such as running or long jumping.
  • Once an athlete's body leaves the ground, they cannot change their horizontal velocity or horizontal path of motion.
• Zero Net Force (Application of Newton's First Law):
  • Applies when no external forces act on an object or when the net sum of external forces equals zero ($\Sigma F = 0$).
  • Static Example: Holding a $10\,\text{lb}$ dumbbell stationary in hand.
  • Involved forces: Downward gravitational weight force ($-W$) and upward reaction force from the hand ($R$).
  • Net force calculation:       $\Sigma F = R + (-W) = 0\,\text{lb}$
  • Constant Velocity Example: Raising a $10\,\text{lb}$ dumbbell at a constant upward velocity.
  • Force exerted against the dumbbell to maintain constant upward movement equals an upward force of $10\,\text{lb}$, yielding a net force of zero.

Newton's Second Law of Motion and Projectile Kinematics

• Newton's Second Law of Motion (Law of Acceleration):
  • Verbatim Definition: If a net external force is exerted on an object, the object will accelerate in the direction of the net external force and its acceleration will be proportional to the net external force and inversely proportional to its mass.
  • Mathematical Formulation ("Cause and Effect"):     $\Sigma F = m \mathbf{a}$
  • Forces cause acceleration.
  • Sample Problem:
  • A $2\,\text{kg}$ ball is kicked with a force of $10\,\text{N}$ (where $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$).
  • Calculation:       $\Sigma F = m \mathbf{a}$$10\,\text{N} = (2\,\text{kg}) \mathbf{a}$$\mathbf{a} = 5\,\text{m/s}^2$
• Vertical Motion of a Projectile:
  • Governed by Newton's Second Law when an object is thrown straight up in the air (zero horizontal motion):     $\Sigma F_y = m a_y$$W = m a_y$
  • Kinematic behavior: The ball slows down while rising in the air and speeds up while falling.
  • Acceleration characteristics:
  • Downward vertical acceleration does NOT change during flight.
  • The ball slows down by the same amount over equal time intervals while rising, and speeds up by the same amount over equal time intervals while falling.
  • Vertical acceleration is constant and directed downward at $9.81\,\text{m/s}^2$.
  • Object weight does NOT affect vertical acceleration.
• Combined Horizontal and Vertical Motion of a Projectile:
  • When an object moves up and out simultaneously, it experiences a constant horizontal velocity alongside a constant vertical acceleration ($9.81\,\text{m/s}^2$ downward).
  • Vertical velocity decreases by $9.81\,\text{m/s}$ for each second of upward flight and increases by $9.81\,\text{m/s}$ for each second of downward flight.

Newton's Third Law of Motion and Interactions

• Newton's Third Law of Motion (Law of Action-Reaction):
  • Verbatim Definition: For every action there is an equal but opposite reaction.
  • Interacting forces are equal in magnitude and opposite in direction.
  • Effect of Body Mass:
  • The object with the smaller mass experiences a greater acceleration or impact effect from the interaction.
  • Practical Example: A $250\,\text{lb}$ linebacker hits a $175\,\text{lb}$ wide receiver in football. Each player exerts an equal magnitude of force on the other, but the impact produces a far greater change in acceleration and trajectory on the lighter $175\,\text{lb}$ wide receiver.
  • Head-on Collision Calculation:
  • A $90\,\text{kg}$ hockey player collides head-on with an $80\,\text{kg}$ player.
  • If the first player exerts a force of $450\,\text{N}$ on the second player, the second player exerts a force of $-450\,\text{N}$ on the first player (equal magnitude of $450\,\text{N}$ in the opposite direction).

Impulse, Linear Momentum, and Applications in Sport

• Linear Momentum ($M$):
  • Verbatim Definition: Quantity of motion, measured as the product of a body's mass and its velocity.     $M = m \mathbf{v}$
  • Static objects have zero velocity, resulting in zero momentum.
  • Linear momentum changes require changing either body mass or velocity.
  • In sports, body or implement mass remains constant during movement, meaning momentum changes occur through changes in velocity.
  • Velocity changes require imposing an external force over a duration of time.
• Impulse:
  • Verbatim Definition: Impulse produced by a net force acting over some duration of time causes a change in momentum.
  • Equation:     $\Sigma F \Delta t = m(V_f - V_i)$
  • $\Sigma F$ = average net force
  • $\Delta t$ = change in time / duration of force application
  • $m$ = mass of the object
  • $V_f$ = final velocity
  • $V_i$ = initial velocity
• Applying Impulse to Increase Momentum:
  • Generating a large change in velocity requires a large average net force acting over an extended time duration.
  • In sports where the goal is to impart high velocity to an object (e.g., baseball pitch, golf swing), athletes apply a maximum force against the object for as long a duration as possible.
• Applying Impulse to Decrease Momentum:
  • When an object has a high initial velocity that must be reduced to zero ($V_f = 0$), impulse is used to absorb momentum.
  • Jump Landing Mechanics:
  • Flexing knees, hips, and ankles upon landing increases the time component ($\Delta t$).
  • Because the total change in momentum $m(V_f - V_i)$ is identical regardless of landing style, increasing $\Delta t$ reduces the average impact force ($\Sigma F$).
  • Landing straight-legged results in a very short impact time ($\Delta t$), forcing average impact force ($\Sigma F$) to increase dramatically, thereby increasing joint strain and injury risk.
  • Protective athletic equipment (e.g., football padding, hockey padding, gymnastics landing mats, high jump mats) functions by increasing force duration ($\Delta t$) to decrease average impact force ($\Sigma F$).
• Impulse-Momentum Calculation:
  • A sled race begins with two crew members pushing a $90\,\text{kg}$ sled initially at rest ($V_i = 0\,\text{m/s}$) with an average force of $100\,\text{N}$ in the direction of motion for $7\,\text{s}$ before jumping in. Neglecting friction, calculate the final speed.
  • Given: $F = 100\,\text{N}$, $\Delta t = 7\,\text{s}$, $m = 90\,\text{kg}$, $V_i = 0\,\text{m/s}$.
  • Calculation:     $\Sigma F \Delta t = m(V_f - V_i)$$(100\,\text{N})(7\,\text{s}) = (90\,\text{kg})(V_f - 0\,\text{m/s})$$700\,\text{N}\cdot\text{s} = (90\,\text{kg}) V_f$$V_f = 7.78\,\text{m/s}$

Collisions, Impact Mechanics, and Coefficient of Restitution

• Impact Definition: Collision of two bodies characterized by the exchange of a large force during a small time interval.
• The post-impact behavior of colliding bodies depends on their collective momentum and relative elasticity.
• In head-on collisions between two objects, total combined linear momentum is conserved.
• Classification of Collisions:
  • Perfectly Elastic Collisions:
  • Relative velocities of the two bodies after impact equal their relative velocities before impact.
  • No loss of system kinetic energy occurs.
  • Examples: Impact of a superball with a hard concrete surface, coin collision interactions.
  • Perfectly Inelastic (Plastic) Collisions:
  • At least one of the contacting bodies deforms permanently without regaining its original shape, and the bodies do not separate post-collision (they move together as a single unit).
  • Example: Dropping modeling clay onto a surface.
  • Most real-world physical impacts fall between perfectly elastic and perfectly plastic extremes.
• Sample Elastic Collision Calculation:
  • A $2.5\,\text{g}$ penny ($A$) moving at $2\,\text{m/s}$ collides with a stationary penny ($B$) of mass $2.5\,\text{g}$. Penny $A$ comes to an immediate stop upon contact. Calculate the post-collision velocity of penny $B$.
  • Formula:     $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
  • Given: $m_1 = 2.5\,\text{g}$, $u_1 = 2\,\text{m/s}$, $m_2 = 2.5\,\text{g}$, $u_2 = 0\,\text{m/s}$, $v_1 = 0\,\text{m/s}$.
  • Calculation:     $(2.5\,\text{g})(2\,\text{m/s}) + (2.5\,\text{g})(0\,\text{m/s}) = (2.5\,\text{g})(0\,\text{m/s}) + (2.5\,\text{g}) v_2$$5\,\text{g}\cdot\text{m/s} - 0 = (2.5\,\text{g}) v_2$$5\,\text{g}\cdot\text{m/s} = (2.5\,\text{g}) v_2$$v_2 = 2\,\text{m/s}$
• Sample Inelastic (Plastic) Collision Calculation:
  • A $90\,\text{kg}$ hockey player moving at $6\,\text{m/s}$ collides head-on with an $80\,\text{kg}$ player moving at $7\,\text{m/s}$ in the opposite direction ($-7\,\text{m/s}$). The two players entangle and continue traveling together as a unit. Find their combined post-collision velocity.
  • Formula:     $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$
  • Given: $m_1 = 90\,\text{kg}$, $v_1 = 6\,\text{m/s}$, $m_2 = 80\,\text{kg}$, $v_2 = -7\,\text{m/s}$.
  • Calculation:     $(90\,\text{kg})(6\,\text{m/s}) + (80\,\text{kg})(-7\,\text{m/s}) = (90\,\text{kg} + 80\,\text{kg}) v$$540\,\text{kg}\cdot\text{m/s} - 560\,\text{kg}\cdot\text{m/s} = (170\,\text{kg}) v$$-20\,\text{kg}\cdot\text{m/s} = (170\,\text{kg}) v$$v = -0.118\,\text{m/s}$
  • Result: The entangled players move together at $0.118\,\text{m/s}$ in the original direction of the $80\,\text{kg}$ player.
• Coefficient of Restitution ($e$):
  • Definition: Value ranging from $0$ to $1$ that describes the relative elasticity of an impact.
  • Values closer to $1$ indicate greater elasticity.
  • Values closer to $0$ indicate greater inelasticity/plasticity.
  • Newton's Law of Impact:     $e = \frac{\text{relative velocity after impact}}{\text{relative velocity before impact}} = \frac{V_1 - V_2}{U_1 - U_2}$
  • $U_1, U_2$ = pre-impact velocities of bodies 1 and 2
  • $V_1, V_2$ = post-impact velocities of bodies 1 and 2
  • Impact between a moving object and a stationary object/surface:     $e = \sqrt{\frac{h_b}{h_d}}$
  • $h_b$ = height to which the ball bounces
  • $h_d$ = height from which the ball is dropped
  • Interaction Property: The coefficient $e$ describes the interaction between two contacting bodies; it is NOT descriptive of any single isolated object or surface. Increasing collision velocity increases $e$
• Coefficient of Restitution Calculation:   
  • A basketball is dropped from a height of $2\,\text{m}$ onto a gymnasium floor. If the coefficient of restitution between the ball and floor is $0.9$, calculate the bounce height.
  • Given: $h_d = 2\,\text{m}$, $e = 0.9$.
  • Calculation:     $e = \sqrt{\frac{h_b}{h_d}}$$0.9 = \sqrt{\frac{h_b}{2\,\text{m}}}$$(0.9)^2 = \frac{h_b}{2\,\text{m}}$$0.81 = \frac{h_b}{2\,\text{m}}$$h_b = 1.62\,\text{m}$

Mechanical Work and Muscle Contraction Mechanics

• Mechanical Work Definition: Force applied against a resistance, multiplied by the displacement of the resistance in the direction of that force.   $W = F D$
  • Units: Newton-meters ($\text{N}\cdot\text{m}$) or Joules ($\text{J}$).
• Work Vector Properties and Signs:
  • Positive Work ($+$): Done when force displacement is in the same direction as the applied force.
  • Examples: Pitcher throwing a baseball, weightlifter raising a weight.
  • Anatomical Equivalent: Concentric muscle contraction.
  • Requires higher caloric expenditure than performing an equal amount of negative work.
  • Negative Work ($-$): Done when force displacement is in the direction opposite to the applied force.
  • Examples: First baseman catching a ball, weightlifter lowering a barbell.
  • Anatomical Equivalent: Eccentric muscle contraction.
  • Zero Work ($0$):
  • Not all muscle contractions produce mechanical work.
  • Anatomical Equivalent: Isometric muscle contraction (muscle contracts, but length remains unchanged, yielding zero displacement $D = 0$
• Required Variables to Determine Mechanical Work:
  • Average force exerted on the object
  • Direction of applied force
  • Displacement of force application
• Bench Press Calculation:
  • A lifter begins a lift with arms extended holding a $1000\,\text{N}$ barbell at $75\,\text{cm}$ ($0.75\,\text{m}$) above his chest. He lowers the weight to $5\,\text{cm}$ ($0.05\,\text{m}$) above his chest and raises it back to full extension.
  • Lowering Phase Calculation:     $F = 1000\,\text{N}$$D = y_f - y_i = 0.05\,\text{m} - 0.75\,\text{m} = -0.70\,\text{m}$$W_{\text{lower}} = (1000\,\text{N})(-0.70\,\text{m}) = -700\,\text{J}\quad\text{(negative work)}$
  • Raising Phase Calculation:     $F = 1000\,\text{N}$$D = y_f - y_i = 0.75\,\text{m} - 0.05\,\text{m} = 0.70\,\text{m}$$W_{\text{raise}} = (1000\,\text{N})(0.70\,\text{m}) = 700\,\text{J}\quad\text{(positive work)}$
  • Complete Repetition Calculation:
  • Total displacement is zero.     $W_{\text{total}} = W_{\text{lower}} + W_{\text{raise}} = -700\,\text{J} + 700\,\text{J} = 0\,\text{J}$

Mechanical Power and Athletic Performance

• Mechanical Power Definition: Rate of work production, measuring how quickly or slowly work is produced.
• Mathematical Formulas:   $P = \frac{W}{\Delta t}$$P = F v$
  • Units: Watts ($\text{W}$) = Joules per second ($\text{J/s}$).
• Stair Climbing Power Calculation:
  • A $580\,\text{N}$ person runs up a flight of 30 stairs, each step having a riser height of $25\,\text{cm}$ ($0.25\,\text{m}$), in $15\,\text{s}$.
  • Given: $F = 580\,\text{N}$, Total Height $H = 30 \times 0.25\,\text{m} = 7.5\,\text{m}$, $t = 15\,\text{s}$.
  • Work Calculation:     $W = F D = (580\,\text{N})(7.5\,\text{m}) = 4350\,\text{J}$
  • Power Calculation:     $P = \frac{W}{t} = \frac{4350\,\text{J}}{15\,\text{s}} = 290\,\text{W}$
• Power in Sports Performance:
  • In sports such as Olympic weightlifting, football, hockey, and track & field, high power output is far more critical than pure weight-room strength.
  • Force-Velocity Dynamics of Muscle:
  • Increasing muscle contraction velocity decreases maximal contraction force (slow contractions produce higher maximum force).
  • Mechanical Power equation: $P = F v$
  • Maximal muscle power output occurs at a contraction velocity approximately $\frac{1}{2}$ of the muscle's maximal contraction velocity.
  • Running Stride Kinematics: Increasing stride frequency, stride length, or both increases running velocity, increasing power. Because force drops at high contraction speeds, optimal stride length and rate are intermediate rather than maximum extremes.
• Sustainable Power and Activity Duration:   
  • Maximal sustainable power output decreases as event duration increases.
  • Sprinters: Maintain high power output over short durations ($0–60\,\text{seconds}$).
  • Middle-Distance Runners: Maintain moderate power output over intermediate durations ($1–7\,\text{minutes}$).
  • Marathon Runners: Maintain low power output over extended durations ($2–4\,\text{hours}$).

Energy Classification and Principles of Conservation

• Energy Definition: Capacity or ability to do mechanical work.
• Kinetic Energy ($KE$):
  • Definition: Energy possessed by an object due to its motion.
  • Equation:     $KE = \frac{1}{2} m v^2$
  • Units: Newton-meters ($\text{N}\cdot\text{m}$) or Joules ($\text{J}$).
  • Mass Effect: Heavier objects possess higher kinetic energy.
  • Velocity Effect: Because velocity is squared, increases in velocity produce dramatic exponential increases in kinetic energy.
  • Baseball Calculation:
  • Mass $m = 0.145\,\text{kg}$, thrown velocity $v = 35.8\,\text{m/s}$ ($80\,\text{mph}$).
  • Calculation:       $KE = \frac{1}{2} (0.145\,\text{kg}) (35.8\,\text{m/s})^2 = 92.9\,\text{J}$
• Potential Energy ($PE$):
  • Definition: Energy possessed by an object due to its position (stored energy).
  • Gravitational Potential Energy ($PE$):
  • Dependent on body weight and height above a baseline reference point.
  • Equation:       $PE = w h = m g h$
    • $g = 9.81\,\text{m/s}^2$
    • Units: Newton-meters ($\text{N}\cdot\text{m}$) or Joules ($\text{J}$).
  • Diver Calculation:
    • A $70\,\text{kg}$ diver stands on a diving board $20\,\text{m}$ above pool surface.
    • Calculation:         $PE = m g h = (70\,\text{kg})(9.81\,\text{m/s}^2)(20\,\text{m}) = 13734\,\text{J}$
  • Practical Application: Increasing hammer mass or strike height above a nail increases gravitational potential energy, increasing nail penetration depth.
  • Strain or Elastic Potential Energy ($SE$):
  • Definition: Energy stored due to deformation of an elastic object.
  • Equation:       $SE = \frac{1}{2} k x^2$
    • $k$ = stiffness or spring constant of the material
    • $x$ = distance over which the object is deformed
  • Pole Vault Application: Vaulting poles bend to store strain energy ($SE$), releasing it as kinetic energy ($KE$) when unbending to propel the athlete upward. Greater pole bend stores greater strain energy.
  • Anatomical Examples: Muscle tendons during contraction, arches of the foot during walking or running.
  • Material Properties: Stiffer materials store greater strain energy when deformed.
• Conservation of Mechanical Energy:   
  • Vertically Tossed Ball Energy Dynamics:
  • As a ball gains height ($h$), gravitational potential energy increases ($PE = mgh$).
  • As height increases, vertical velocity ($v$) decreases due to gravity ($g$), reducing kinetic energy ($KE = \frac{1}{2}mv^2$).
  • At peak apex height, potential energy reaches maximum while velocity and kinetic energy equal zero.
  • As the ball falls downward, kinetic energy progressively increases as velocity increases, while potential energy decreases as height decreases.
  • Law Statement: When gravity is the only acting external force, total mechanical energy ($C$) remains constant.     $C = PE + KE$
  • Constant Mechanical Energy Dropped Ball Calculation:
  • A $2\,\text{kg}$ ball is dropped from a height of $1.5\,\text{m}$. Calculate velocity immediately prior to floor impact.
  • Total Mechanical Energy ($C$) at initial release height ($1.5\,\text{m}$):       $C = PE + KE$$C = m g h + \frac{1}{2} m v^2$$C = (2\,\text{kg})(9.81\,\text{m/s}^2)(1.5\,\text{m}) + \frac{1}{2}(2\,\text{kg})(0\,\text{m/s})^2$$C = 29.43\,\text{J} + 0 = 29.43\,\text{J}$
  • Velocity immediately before impact ($h = 0\,\text{m}$):       $C = PE + KE$$29.43\,\text{J} = (2\,\text{kg})(9.81\,\text{m/s}^2)(0\,\text{m}) + \frac{1}{2}(2\,\text{kg}) v^2$$29.43\,\text{J} = 0 + (1\,\text{kg}) v^2$$v^2 = 29.43\,\text{m}^2/\text{s}^2$$v = 5.42\,\text{m/s}$

The Work-Energy Relationship

• Fundamental Principle: Work and energy are directly proportional. Mechanical work done by external forces (excluding gravity) equals the change in total energy of the system.
• Governing Equation:   $W = \Delta KE + \Delta PE + \Delta TE$
  • $\Delta TE$ = change in thermal energy (heat)   $W = (KE_f - KE_i) + (PE_f - PE_i) + (TE_f - TE_i)$
• Doing Work to Increase Energy:
  • Athletic movement performance requires changing velocity, which directly corresponds to changing kinetic energy ($KE$).
  • Achieving a large change in kinetic energy requires applying a large force over a long distance.
  • Aligns directly with impulse-momentum principles ($\Sigma F \Delta t = m(V_f - V_i)$).
  • Increasing work increases change in kinetic energy, which increases final object velocity.
• Doing Work to Decrease Energy:
  • When catching a thrown ball, negative work must be done to reduce or absorb the ball's kinetic energy down to zero.
• Catching Baseball Calculation:
  • Calculate work required to catch a $1.3\,\text{kg}$ baseball moving at $40\,\text{m/s}$.
  • Given: $m = 1.3\,\text{kg}$, $v_i = 40\,\text{m/s}$, $v_f = 0\,\text{m/s}$. Assume $\Delta PE = 0$ and \Delta TE = 0$.\n * Calculation:\n    W = \Delta KE + \Delta PE + \Delta TE\n    W = (KE_f - KE_i) + 0 + 0\n    W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2\n    W = 0 - \frac{1}{2}(1.3\,\text{kg})(40\,\text{m/s})^2\n    W = 0 - (0.65\,\text{kg})(1600\,\text{m}^2/\text{s}^2)\n    W = -1040\,\text{J}$$