9.4 Work-Energy Principle, Power, and Efficiency

9.4 Work-Energy Principle, Power, and Efficiency

The work-energy method trades time for distance: it answers "how fast is it moving after traveling this far" without ever integrating an acceleration. NCEES lists work, energy, and power as a Dynamics sub-topic, and the energy route is almost always the faster of the two available approaches.

Principle of Work and Energy

The Work-Energy Principle is scalar-based and directly relates force, displacement, and velocity without requiring calculation of acceleration.

$$T_1 + U_{1-2} = T_2$$

Kinetic Energy ($T$)

• Particle Translation: $T = \frac{1}{2} m v^2$
• Rigid Body Fixed-Axis Rotation: $T = \frac{1}{2} I_O \omega^2$
• Rigid Body General Planar Motion:

$$T = \frac{1}{2} m v_G^2 + \frac{1}{2} I_G \omega^2$$

Work Done by Forces ($U_{1-2}$)

• Variable Force: $U_{1-2} = \int_{s_1}^{s_2} F \cos \theta ds$
• Constant Force: $U_{1-2} = F \cos \theta (s_2 - s_1)$
• Weight (Gravity): $U_{1-2} = -m g \Delta y$
• Linear Spring: $U_{1-2} = -\frac{1}{2} k (x_2^2 - x_1^2)$
• Constant Moment / Couple: $U_{1-2} = M (\theta_2 - \theta_1)$

Conservation of Mechanical Energy

When only conservative forces (gravity, springs) perform work, total mechanical energy is conserved:

$$T_1 + V_{g1} + V_{e1} = T_2 + V_{g2} + V_{e2}$$

where gravitational potential energy $V_g = m g h$ and elastic potential energy $V_e = \frac{1}{2} k x^2$.

Power

Power is the rate of doing work:

$$P = \frac{dW}{dt} = \vec{F}\cdot\vec{v} = Fv\cos\theta \qquad\text{(translation)}$$ $$P = T\omega \qquad\text{(rotation, with }\omega\text{ in rad/s)}$$

The rotational power trap: $P = T\omega$ requires $\omega$ in rad/s, not rev/min. Feeding rpm directly in overstates power by $2\pi/60 = 9.55$. A useful US Customary shortcut that bakes in the conversion: $\text{hp} = \frac{T\ [\text{ft}\cdot\text{lbf}] \times N\ [\text{rpm}]}{5{,}252}$

Efficiency

$$\eta = \frac{\text{useful output}}{\text{total input}}$$

For components in series, efficiencies multiply:

$$\eta_{\text{overall}} = \eta_1 \times \eta_2 \times \cdots \times \eta_n$$

This compounding is why long drive trains lose so much: three stages at 90% each deliver $0.9^3 = 72.9\%$, not 90%.

Worked Example: Pump Motor Sizing Through a Drive Train

A pump must deliver 45 kW of hydraulic power. It is driven through a belt drive ($\eta = 0.94$) by a motor ($\eta = 0.91$), and the pump itself is 78% efficient. Find the electrical input power and the motor torque at 1{,}750 rpm.

Work backwards from the useful output. The pump's shaft input:

$$P_{\text{pump shaft}} = \frac{45}{0.78} = 57.7\ \text{kW}$$

Through the belt:

$$P_{\text{motor shaft}} = \frac{57.7}{0.94} = 61.4\ \text{kW}$$

Electrical input to the motor:

$$P_{\text{electrical}} = \frac{61.4}{0.91} = \boxed{67.5\ \text{kW}}$$

Check with the compounded efficiency: $\eta_{\text{overall}} = 0.78(0.94)(0.91) = 0.667$, and $45/0.667 = 67.5$ kW ✓

Motor torque:

$$\omega = \frac{2\pi(1{,}750)}{60} = 183.3\ \text{rad/s}$$ $$T = \frac{P}{\omega} = \frac{61{,}400}{183.3} = \boxed{335\ \text{N}\cdot\text{m}}$$

Note which power drives the torque: the motor's shaft output (61.4 kW), not its electrical input. The 6.1 kW difference is dissipated as heat in the motor windings and never appears as mechanical torque. Two-thirds of the electrical input reaches the water, and the delivered fraction would drop to 55% if each stage lost just five more points of efficiency.

When to Use Energy vs. Newton's Second Law

Choosing correctly is worth more time on this exam than computing faster. A problem giving a distance and asking for a velocity is an energy problem; reaching for $F = ma$ forces you to integrate an acceleration you do not need.