PHYS 218 Chapter 6

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Work

Work of a force over an object moving from $r_1$ to $r_2$

$W=\underset{r_1}{\overset{r_2}{\int }}F\cdot dl$

• $F$ is force
• $\cdot$ is the dot product
• $dl$ is the displacement vector</math>

If $F$ is constant (in magnitude, direction, and time), then $W=F\cdot (r_2-r_1)$

Units

Work is measured in Joules [J] (J = N × m = $\frac{kg\times m^2}{s^2}$ )

Example

A train moves 100 meters along a straight track with a force of 1000 N applied at an angle of 60°

Since force is constant, $W=F\cdot \mathrm{\Delta }r=1000\cos 60^{\circ }\cdot 100=50000\text{J}$

Work by Net Forces

Given forces $F_1$, $F_2$, and $F_3$,

$\begin{array}{rlr}W& =\underset{r_1}{\overset{r_2}{\int }}\sum F\cdot dl& (\text{where}dl\text{is change in displacement})\\ & =\underset{r_1}{\overset{r_2}{\int }}ma\cdot dl\\ & =\underset{r_1}{\overset{r_2}{\int }}m\frac{dv}{dt}\cdot dl\\ & =\underset{r_1}{\overset{r_2}{\int }}mdv\cdot \frac{dl}{dt}\\ & =\underset{r_1}{\overset{r_2}{\int }}mvdv\\ & =\frac{m{v_2}^2}{2}-\frac{m{v_1}^2}{2}\end{array}$

Net work is related to change in velocities.

Kinetic Energy

$K=\frac{m{v}^2}{2}$ (J)

Work Energy Theorem

Work of all external forces (J) = change in kinetic energy (J)
$W=K_2-K_1$ (J)

Example

Object:

• mass = 2 kg
• initial velocity = 10 m/s
• presence of gravity

What is work of gravitational force

1. object from bottom to top
2. object from top to bottom
3. object during complete journey

1. $\begin{array}{rlrl}W& =K(t_T)-K(t_0)& =0-\frac{1}{2}\times 2\times 100& =-100\end{array}$

2. $W=100-0$

3. $W=100-100=0$

Example 2

Object:

• mass = 10 kg
• initial velocity = 2 m/s
• Force applied = 110 N

Find velocity when displacement is at 20 m.

$\begin{array}{rl}W& =K_2-K_1\\ & =\frac{m{v_2}^2}{2}-\frac{m{v_1}^2}{2}\\ & =20\\ 2200& =\frac{m{v_2}^2}{2}-20\\ & =\dots \end{array}$

Springs

An uncompressed spring's length is $l_0$ or natural length

In small increments, the force that the spring exerts on the block is $-k(x-l_0)$, where $k$ is the spring constant [N/m]

• $x>l_0\to F<0$
• $x<l_0\to F>0$

Example

An object of mass $m$ is pushing against a spring. When let go, the spring pushes the object to a known velocity and returns to its natural length

Find the work done by the spring

$\begin{array}{rl}W& =\underset{r_1=l_0-5}{\overset{r_2=l_0}{\int }}-k(x-l_0)dx\\ & =-k\underset{l_0-5}{\overset{l_0}{\int }}x-l_{0}dx\\ & =-k(\frac{{l_0}^2}{2}-\frac{(l_0-5{)}^2}{2}-l_0(l_0-(l_0-5)))\\ & =\dots \frac{k\mathrm{\Delta }{x}^2}{2}=\frac{25k}{2}& \text{energy stored in spring when compressed}\end{array}$

Power

Rate of change of Work over time measured in Joules per second or watts [ J/s = W ]

$\begin{array}{rl}P& =\frac{dW}{dt}\\ & =F\cdot (v_2-v_1)& \text{if force is constant}\end{array}$

For example, a 100 Watt light bulb is converting 100 J of work into light each second.

Note: Energy through power: kilowatt hour (energy in units of joules)