10-24-25 Energy Dissipation, Power

Energy Dissipation

  • Kinetic friction involves a large number of atomic collisions in random directions

  • Having collisions in random directions causes the kinetic energy of the object to spread out (dissipated) into many directions

Air Resistance

  • Air resistance: Friction-like force felt by an object moving through air

  • If the speed through the air is low enough that turbulence is not created, this force is given by the following formula, where:

    • CC is a constant determined by how aerodynamic the object is

    • ρ\rho is the density of the air

    • AA is the cross-sectional area of the object

F→air=approx−12CρA∣v→∣2v‾\overrightarrow F_{air} =_{approx} - \frac 12 C\rho A |\overrightarrow v|² \overline v

Term)inal Speed

  • Terminal Speed: Maximum speed of a falling object

  • Terminal speed occurs when the force of air resistance is large enough to balance the gravitational force

Practice:

A single falling coffee filter quickly reaches a terminal speeds of 0.5m/s0.5 m/s. At this speed, what do we know about the air resistance force?

A)A) F_{air} > F_{grav}

B)B) Fair=FgravF_{air} = F_{grav}

C)C) F_{air} < F_{grav}

D)D) Not enough information

The terminal speed has been reached. This means that air resistance is balanced with gravity.

The correct answer is B

Practice:

A coffee filter with a mass of 1.2 grams is dropped from rest at a height of 3 m above the ground. The filter reaches the ground with a speed of 0.5m/s0.5 m/s. How much internal energy did the air + coffee filter system gain while the coffee filter was falling?

A)A) 0.03528J0.03528J

B)B) 0.03513J0.03513J

C)C) 0.01500J0.01500J

D)D) Not enough information

Gather)

mfilter=1.2gm_{filter} = 1.2g

r_i = <0,3,0>m

v_i = <0, 0.5, 0> m/s

r_f = <0,0,0>m

v_f = <0,0,0> m/s

ΔEint=??\Delta E_{int} = ??

Organize)

System: Air, Coffee filter

  • ⇒ΔEsys=ΔErest+ΔK+ΔU+ΔEint\Rightarrow \Delta E_{sys} = \Delta E_{rest} + \Delta K + \Delta U + \Delta E_{int}

    • ΔErest\Delta E_{rest} and ΔU\Delta U can be ignored since they are not changing

  • ΔEsys=(ΔKfilter+ΔKair)+ΔEint\Delta E_{sys} = (\Delta K_{filter} + \Delta K_{air}) + \Delta E_{int}

    • ΔKair\Delta K_{air} can be ignored since, well, you know, ummmm, air doesn’t really change

Surroundings: Earth

  • ⇒∑\Rightarrow \sum Transfers =Wsur+Q+EEM+…= W_{sur} + Q + E_{EM} + …

    • Q,EEM,Q, E_{EM}, and …… can be ingored

  • ∑\sum Transfers =F→g⋅Δr→= \overrightarrow F_g \cdot \Delta \overrightarrow r

Analyze)

  • ΔEsys=∑\Delta E_{sys} = \sum Transfers ⇒Kfilter+ΔEint=F→g⋅r→\Rightarrow K_{filter} + \Delta E_{int} = \overrightarrow F_g \cdot \overrightarrow r

  • ⇒ΔEint=F→g⋅Δr→−ΔKfilter=∣mg∣∣Δr→∣cos⁡(0°)−(12mvf2−12mvi2)\Rightarrow \Delta E_{int} = \overrightarrow F_g \cdot \Delta \overrightarrow r - \Delta K_{filter} = |mg| |\Delta \overrightarrow r| \cos (0\degree) - (\frac 12 mv_f² - \frac 12 mv_i²)

    • 12mvi2\frac 12 mv_i² can be crossed out

  • ⇒ΔEint=0.0012kg⋅9.8Nkg⋅3m−12⋅0.0012kg⋅(0.5ms)2\Rightarrow \Delta E_{int} = 0.0012 kg \cdot 9.8 \frac N{kg} \cdot 3 m - \frac 12 \cdot 0.0012 kg \cdot (0.5 \frac ms)²

  • ⇒ΔEint=0.03513J\Rightarrow \Delta E_{int} = 0.03513 J

The answer is B

Spring-Mass System with Firction

  • A spring-mass system that experiences viscous friction (where force is proportional to velocity) will have an amplitude that decreases over time.

f→v=−bv→\overrightarrow f_v = -b\overrightarrow v

F→s=−kssL→\overrightarrow F_s = -k_s s\overrightarrow L

F→net=−kssL‾−bv→\overrightarrow F_{net} = -k_s s\overline L -b\overrightarrow v

$$\overrightarrow F_{net} = m \frac {d²x}{dt²} \Rightarrow Ae^{-(\frac b{2m}t} \cos(\omega t)