Unit 5 Summary

Work and Power

work = a transfer of power
work = (force)(distance)   --> measured in joules (J)
*but remember that force and distance must be parallel for work to be done

For example, if a man was walking down the street holding a dog, no work would be done on the dog because the man's distance is moving forward and the dog's gravity is downward.
... But if the man was picking up the dog from the ground, there would be work on the dog, because the man is moving upward thus making the force and the distance parallel.

There is work being done, because the man's weight is 600N and the stairs have a vertical height of 4m
--> the force and distance are parallel
*the slanted height of the stairs DOES NOT matter because it isn't parallel with the force


Example problem: There are two people who weigh 500N. One man walked up the stairs, and the other man took the elevator. Which person did gravity do more work on?
-->

*There is an exception to the force and distance parallel rule.

Work can still be done if something is slanted, because gravity is still parallel to the y-component, so it doesn't have to be completely parallel to the actual distance (velocity in this case).





power = how quickly work is done
power = work/time   --> measured in watts
*If you halve the time, you double the work

Work and Kinetic Energy

kinetic energy - energy of movement

kinetic energy = .5mv^2   --> measured in joules (J)

work = ∆ kinetic energy

*if you have mass & velocity, you have the potential to do work

Say an 8g bike was moving at 10m/s and then 10 seconds later was moving at 20m/s. What was the 

∆ kinetic energy?

KE initial = .5mv^2                 KE final = .5mv^2                  

                  = .5(8)(10)^2                          = .5(8)(20)^2
                  = 4(100)                                  = 4(400)
                  = 400 J                                    = 1,600J

∆ kinetic energy = KE final - KE initial
                             = 1,600 J - 400J
                             = 1,200J  <--------- ∆ KE

work = ∆ KE so since there was 4x as much KE in KE new than in KE original, we know there was 4x as much work in KE new than in KE original.

*remember when you're solving for work that kg IS NOT a force, so you must convert the kg into N using w=mg because weight IS a force.
Go two decimal places to the left when converting from kg --> N
Go two decimal places to the right when converting from N --> kg

Potential Energy

potential energy = mgh --> (mass)(gravity)(height)
*movement IS NOT necessary to have potential energy
*the amount of potential energy is affected by the height (energy of position), so if something has height it has potential energy.

PE      = ∆KE        = work
(mgh) = (.5mv^2) = (F)(d)

The potential energy = the kinetic energy
         (potential)                     (actual)





*The Law of Conservation of Energy says that, "the total amount of energy in a system remains constant, or is conserved. Energy can't be created nor destroyed, but it can change form." Just like in this problem PE = KE so the energy went from potential energy to kinetic energy.

PE = mgh 
*if you double the height, you double the PE
*if you double the mass, you double the PE
*if you double the height & mass, you quadruple the PE
*gravity always remains constant .98 or 10

Machines

These are three different machines.
left = ramp
middle = pulley
right = lever

The purpose of machines is to decrease the amount of force needed, thus making the job easier.

The work put in = the work that comes out
work in = work out 
  (F)(d)  =  (F)(d)

In terms of work and energy, how do airbags keep you safe?
Airbags keep you safe, because you go from moving to not moving no matter how you're stopped, because ∆KE is the same with or without the airbag because (KE = .5mv^2) and
∆KE = KE final - KE initial. Since the ∆KE is the same, the work is also constant because
(∆KE = work). The airbag increases the distance to stop, and since work has to be the same, the force will decrease with the airbag because (work = Fd), thus having a smaller force.
(small force = small injury = safer)
work = Fd <-- no airbag
work = fD <-- airbag

In terms of work and energy, why do gymnasts use mats?
Gymnasts use mats because they keep them safer. They will go from moving to not moving no matter how they're stopping because ∆KE is the same with or without a mat because (KE = .5mv^2) and
(∆KE = KE final - KE initial). Since the ∆KE is the same, the work is also constant because
(∆KE = work). The mats increase the distance to stop, and since the work has to be the same, the force will decrease with the mats because (work = Fd), thus having a smaller force.
(small force = small injury = safer)
work = Fd <-- no mat
work = fD <-- mat