PHYS 218 Chapter 6

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Work

Work of a force over an object moving from $r_{1}$ to $r_{2}$

$W=\int _{r_{1}}^{r_{2}}F\cdot \ dl$

• $F$ 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 = ${\tfrac {kg\times m^{2}}{s^{2}}}$ )

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, $W=F\cdot \Delta r=1000\cos {60^{\circ }}\cdot 100=50000\ {\mbox{J}}$

Work by Net Forces

Given forces $F_{1}$, $F_{2}$, and $F_{3}$,

${\begin{aligned}W&=\int _{r_{1}}^{r_{2}}\sum {F}\cdot dl&({\mbox{where}}\ dl\ {\mbox{is change in displacement}})\\&=\int _{r_{1}}^{r_{2}}ma\cdot dl\\&=\int _{r_{1}}^{r_{2}}m{\frac {dv}{dt}}\cdot dl\\&=\int _{r_{1}}^{r_{2}}m\,dv\cdot {\frac {dl}{dt}}\\&=\int _{r_{1}}^{r_{2}}mv\ dv\\&={\frac {m{v_{2}}^{2}}{2}}-{\frac {m{v_{1}}^{2}}{2}}\end{aligned}}$

Net work is related to change in velocities.

Kinetic Energy

$K={\frac {mv^{2}}{2}}$ (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. ${\begin{aligned}W&=K(t_{T})-K(t_{0})&=0-{\tfrac {1}{2}}\times 2\times 100&=-100\end{aligned}}$

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.

${\begin{aligned}W&=K_{2}-K_{1}\\&={\frac {m{v_{2}}^{2}}{2}}-{\frac {m{v_{1}}^{2}}{2}}\\&=20\\2200&={\frac {m{v_{2}}^{2}}{2}}-20\\&=\ldots \end{aligned}}$

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]

• $x>l_{0}\rightarrow F<0$
• $x<l_{0}\rightarrow F>0$

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

${\begin{aligned}W&=\int _{r_{1}=l_{0}-5}^{r_{2}=l_{0}}-k(x-l_{0})\ dx\\&=-k\int _{l_{0}-5}^{l_{0}}x-l_{0}\ dx\\&=-k\left({\frac {{l_{0}}^{2}}{2}}-{\frac {(l_{0}-5)^{2}}{2}}-l_{0}(l_{0}-(l_{0}-5))\right)\\&=\ldots {\frac {k\ \Delta x^{2}}{2}}={\frac {25k}{2}}&{\mbox{energy stored in spring when compressed}}\end{aligned}}$

Power

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

${\begin{aligned}P&={\frac {dW}{dt}}\\&=F\cdot (v_{2}-v_{1})&{\mbox{if force is constant}}\end{aligned}}$

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)