1 / 7

Work and Kinetic Energy Work done is defined as Force times the Distance

Work and Kinetic Energy Work done is defined as Force times the Distance. For a Force, F, recall Newton’s Law: F = Ma = Mass X Acceleration Acceleration =change in speed/change in time = ∆V/ ∆t and Speed, V= ∆D/ ∆t for a Distance D and a time t

chelsey
Télécharger la présentation

Work and Kinetic Energy Work done is defined as Force times the Distance

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Work and Kinetic EnergyWork done is defined as Force times the Distance • For a Force, F, recall Newton’s Law: • F = Ma = Mass X Acceleration Acceleration =change in speed/change in time = ∆V/ ∆t and Speed, V= ∆D/ ∆t for a Distance D and a time t …and where ∆ means “change in”

  2. What are the Units of Acceleration? Acceleration Units=Δ Speed/ Δ time • [(L/T)/T]= [L/T2] • Division of fractions: 10/5/2 means 10/5 divided by 2/1 and because this means to invert the divisor and multiply it by the quotient: • Like 10/5 X 1/2 = 1 so with symbols a,b,c…: • a/b/c=(a/b)/c =a/b X 1/c=a /(b X c) or • (L/T/T)=[L/(T X T)]=[L/T2]

  3. What are the Units of Force? • [Force Units]=Mass X Acceleration • =[Mass L/T2] • Like [Kg meter /second2], units of Force. • A Force is applied to an object (Mass, M) so its motion is changed which happens over some time, small or large.

  4. Given the speed, we can know the distance traveled for the time interval over which the work is done. • Since Speed, V, equals Distance, D, divided by time or change in Distance over change in time: • Δ V = ΔD/Δt for example as in Speed = [L/T] like [Miles/Hour] and we can solve for ΔD: • ΔD= Δ V X Δt

  5. When Work is done, the motion changes in response to the force. • For any Force applied to a Mass, the Velocity (Speed), V, must change between the initial and final time. • So we need to find the average of the speed over the (initial to final) time interval, over which the changes happen: Vave=( Vf + Vi ) / 2 and (Df – Di)= Δ V X Δt and if Vi, =0, Di=0 then let D =Df= Δ V X Δt and Vave= Vf / 2 = V/ 2.

  6. Work done is defined as the Force times the Distance • F X D = M X a = M X ∆V/∆t X D • =M X ∆V/∆t Vave ∆t • Remember D= Vave X Δt • ∆t/∆t=1 • F X D =M ∆V Vave = M ∆V V/2 • F X D = M V V/2=½MV2 Work done, F X D = Kinetic Energy, ½MV2

  7. Work done = Kinetic Energy • Kinetic Energy is defined as ½MV2,the energy of movement. • Measured in energy units, Joules=[Mass L2/T2] like [Kg m2/t2]. • Typical Values of Work Hopping Flea 10-7 J Depressing typewriter key 10-2 J Human Heartbeat 0.5 J Lifting 25 lbs through one meter 102 J Hard hit Baseball 103 J Human body resting for one hour 105 J Woman running for one hour 106 J One day of heavy manual labor 107 J Annual Tidal Friction Slowing the Earth 1020 J (http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/TypicalValues.html)

More Related