2.4 The Work-Energy Theorem, Power & Mechanical Advantage

2.4 The Work-Energy Theorem, Power & Mechanical Advantage

The Work-Energy Theorem & Conservation Laws

The work-energy theorem bridges forces and kinematics by stating that the net work performed on an object by all acting forces equals the change in its kinetic energy:

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

Conservative vs. Non-Conservative Forces

Forces in nature fall into two distinct thermodynamic categories:

1. Conservative Forces: Work done is path-independent and depends solely on the initial and final coordinates. A potential energy function can be defined, and the net work done along any closed round-trip path is zero ($\oint \vec{F} \cdot d\vec{r} = 0$). Examples: gravity, electrostatic forces, and ideal spring forces.
2. Non-Conservative (Dissipative) Forces: Work done is path-dependent; mechanical energy is irreversibly converted into thermal energy, sound, or microscopic structural deformation. Examples: kinetic friction, air resistance, and fluid viscous shear stress.

Generalized Energy Conservation

When both conservative and non-conservative forces act on a biological system:

$$W_{\text{nc}} = \Delta E_{\text{mech}} = \Delta KE + \Delta PE$$ $$(KE_i + PE_i) + W_{\text{nc}} = KE_f + PE_f$$

If non-conservative forces are absent ($W_{\text{nc}} = 0$):

$$E_{\text{mech}, i} = E_{\text{mech}, f} \iff KE_i + PE_i = KE_f + PE_f$$

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Power in Biological Systems

Power ($P$) is the rate at which work is performed or energy is converted over time:

$$P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$$

Instantaneous mechanical power delivered by a force $\vec{F}$ moving an object at velocity $\vec{v}$ is:

$$P = \vec{F} \cdot \vec{v} = F v \cos\theta$$

The SI unit of power is the Watt ($1\text{ W} = 1\text{ J/s} = 1\text{ N}\cdot\text{m/s} = 1\text{ kg}\cdot\text{m}^2/\text{s}^3$).

Metabolic and Cardiac Power

• Basal Metabolic Rate (BMR): A resting $70\text{ kg}$ human consumes roughly $80\text{–}100\text{ W}$ of power (metabolizing approximately $2000\text{ kcal/day}$). During vigorous athletic exertion, skeletal muscles can output $> 1000\text{ W}$ of total power.
• Cardiac Hydraulic Power: The rate of work done by the heart to pump cardiac output against systemic vascular resistance is: $$P_{\text{cardiac}} = \Delta P \times Q = \text{MAP} \times \text{CO}$$ where MAP is in Pascals and CO (cardiac output) is in $\text{m}^3/\text{s}$.

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Mechanical Advantage & Simple Machines

A simple machine (lever, inclined plane, pulley, hydraulic lift) redistributes mechanical work: it reduces the required input effort force ($F_{\text{in}}$) by proportionately increasing the distance over which that force must be applied ($d_{\text{in}}$).

Ideal Mechanical Advantage vs. Actual Mechanical Advantage

• Mechanical Advantage ($MA$): Ratio of output load force to input effort force: $$MA = \frac{F_{\text{out}}}{F_{\text{in}}}$$
• Ideal Mechanical Advantage ($IMA$): Ratio of input displacement to output displacement in a frictionless, ideal system: $$IMA = \frac{d_{\text{in}}}{d_{\text{out}}}$$
• Efficiency ($\eta$): Ratio of useful work output to total work input: $$\eta = \frac{W_{\text{out}}}{W_{\text{in}}} = \frac{F_{\text{out}} d_{\text{out}}}{F_{\text{in}} d_{\text{in}}} = \frac{MA}{IMA}$$ In real mechanical systems with friction, $\eta < 1.0$ ($< 100\%$), meaning $MA < IMA$.

Classic Simple Machines

1. Inclined Plane: $$IMA = \frac{L}{h} = \frac{1}{\sin\theta}$$ Pushing a stretcher up a ramp of length $L$ to height $h$ requires an effort force $F = m g \sin\theta$ (neglecting friction), reducing required force by a factor of $L/h$.
2. Pulley Systems: In an ideal block-and-tackle system, $IMA$ equals the number of load-supporting tension strands of rope supporting the movable pulley. If four strands support a patient traction frame, $F_{\text{in}} = \frac{1}{4} F_{\text{load}}$, but the operator must pull $4\text{ m}$ of rope to elevate the limb by $1\text{ m}$.

Summary of Simple Machines and Mechanical Advantage

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Worked Numerical Example: Left Ventricular Stroke Work & Power

Problem: A resting patient has a heart rate of $72\text{ beats/min}$, a stroke volume of $70\text{ mL}$ ($7.0 \times 10^{-5}\text{ m}^3$), and a mean arterial pressure of $100\text{ mmHg}$. Given that $1\text{ mmHg} = 133.3\text{ Pa}$, calculate:

1. The external stroke work ($SW$) performed by the left ventricle during a single systolic contraction.
2. The mechanical hydraulic power output of the left ventricle.

Step 1: Convert MAP to Pascals (SI units)

$$\text{MAP} = 100\text{ mmHg} \times 133.3\text{ Pa/mmHg} = 1.333 \times 10^4\text{ Pa} = 1.333 \times 10^4\text{ N/m}^2$$

Step 2: Calculate Stroke Work ($SW$)

$$SW = \text{MAP} \times \text{SV}$$ $$SW = (1.333 \times 10^4\text{ N/m}^2)(7.0 \times 10^{-5}\text{ m}^3) = 0.9331\text{ J} \approx 0.93\text{ J}$$

Each beat of the left ventricle performs approximately $0.93\text{ Joules}$ of external mechanical work.

Step 3: Calculate Cardiac Output ($CO$) in SI units

$$\text{CO} = \text{HR} \times \text{SV} = \left(\frac{72\text{ beats}}{60\text{ s}}\right) \times (7.0 \times 10^{-5}\text{ m}^3) = (1.2\text{ s}^{-1})(7.0 \times 10^{-5}\text{ m}^3) = 8.4 \times 10^{-5}\text{ m}^3/\text{s}$$

(Note: $8.4 \times 10^{-5}\text{ m}^3/\text{s} = 5.04\text{ L/min}$).

Step 4: Calculate Hydraulic Power Output ($P$)

$$P = \text{MAP} \times \text{CO} = (1.333 \times 10^4\text{ N/m}^2)(8.4 \times 10^{-5}\text{ m}^3/\text{s}) = 1.12\text{ W}$$

The mechanical power delivered by the left ventricle to circulating blood is approximately $1.12\text{ Watts}$. When myocardial metabolic efficiency is factored in (typically $\approx 15\text{–}20\%$), total cardiac chemical energy expenditure is approximately $5\text{–}7\text{ W}$.