Work and Energy

Introduction

The principle of work and energy relates the work done by forces to changes in a particle’s kinetic energy. Rather than solving for acceleration and time, energy methods allow motion to be analyzed using only initial and final states. This is useful when forces vary with position.

Principle of Work and Energy

The total work done by all forces equals the change in kinetic energy. This equation is derived using $\Sigma F = ma$ and $ads = vdv$:

Work of Common Forces

Kinetic energy $= T = \frac{1}{2}mv^{2}$

Gravitational potential energy $= V = mgh$

Variable force: $U = \int_{s_{1}}^{s_{2}}Fds$

Spring potential energy $= F_{s} = \frac{1}{2}kx^{2}$

Constant Force: $U = F(s_{2} - s_{1})cos\theta$

If $\theta = 0{^\circ}$, $U = F\Delta s$

Gravitational potential energy can be positive or negative while kinetic energy is always positive.

Conservation of Energy

A conservative force is one whose work depends only on the initial and final positions, not on the path taken. Common conservative forces are gravity and spring force. Nonconservative forces depend on the path taken, and include friction and drag.

If only conservative forces act:

$T_{1} + V_{1} = T_{2} + V_{2}$

In other words, ${KE}_{1} + {PE}_{1} = {KE}_{2} + {PE}_{2}$

Common applications of conservation:

Spring energy $\rightarrow$ Kinetic energy (spring $\rightarrow$ velocity)

Kinetic energy $\rightarrow$ Potential energy (velocity $\rightarrow$ height)

Potential energy $\rightarrow$ Kinetic energy (height $\rightarrow$ velocity)

Power and Efficiency

Power is the rate at which work is done: $P = \frac{dU}{dt} = \overrightarrow{F} \cdot \overrightarrow{v} = Fvcos\theta$

Efficiency $\eta = \frac{power\ output}{power\ input} = \frac{energy\ output}{energy\ input}$

Example 1:
Find: Height of the roller coaster to achieve a speed of 100 km/hr at point B.

$(\frac{100\ km}{hr})(\frac{1000\ m}{km})(\frac{1\ hr}{3600\ s}) = 27.78\ m/s$

Potential energy $\rightarrow$ kinetic energy:

$mgh = \frac{1}{2}mv^{2}$

$m(9.81\ m/s^{2})(h) = \frac{1}{2}m(27.78\ m/s)^{2}$ $\Rightarrow h = 39.33\ m$

Example 2:
Find the required unstretched length of spring if the spring is compressed 0.2 feet when the 4-lb block slides into it at 9 ft/s. The spring is confined by a plate so that its initial length is 1.5 ft. Neglect any friction or energy loss.

Initially, the spring is compressed $x_{i} = L_{o} - 1.5\ ft$

At the end, the spring is compressed an additional $0.2\ ft$, so $x_{f} = L_{o} - 1.3\ ft$

Kinetic energy + spring potential energy $\rightarrow$ spring potential energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2}$

$\frac{1}{2}(4\ lb/32.2)(9\ ft/s)^{2} + \frac{1}{2}(50\ lb/ft)(L_{o} - 1.5\ ft)^{2} = \frac{1}{2}(50\ lb/ft)(L_{o} - 1.3\ ft)^{2}$

$L_{o} = 1.9\ ft$

Example 3:
Find the power generated by a 150-lb man running up 15-ft high stairs in 4 s.

Work done against gravity: $U = Wh = (150\ lb)(15\ ft) = 2250\ lb \cdot ft$

$P = \frac{dU}{dt} = \frac{2250\ lb \cdot ft}{4\ s} = 562.5\ ft \cdot lb/s \approx 763\ W = 763\ J/s$

Example 4:
Find the work of the force when it displaces 2 m.