2WorkWhen a force is done on an object, energy is transferred to the object.The amount of energy transferred is called work.
3WorkIf you apply a 200.-Newton force to move a box 50.0 meters, how much work have you done?
4Work Symbol: W SI Unit: Joule Equation: W = Fd Force must be in same direction as movement.Work only happens if a force causes movement.
5Is it Work?You push against the wall. The wall stubbornly refuses to move.Is it Work?No! Work is only done if a force causes movement.
6Is it Work? An apple falls from a tree to the ground. Yes! Because the force of gravity causes the apple to move, work is done on the apple.
7Is it Work?You carry a tray above your head while walking across the room at constant speed.Is it work?No! Although you are applying a force to the tray, it isn’t in the same direction that the tray is moving.
8Is it Work? You push a lawnmower across the lawn. Yes! Only the horizontal component of your applied force does work on the lawnmower, however.
9Work or Not? An easy way to tell if work is being done: Is a force acting on an object to cause a displacement?Is the amount of energy an object has (either kinetic or potential) changing?If so, then work is being done!
10Practice ProblemsA 50.0-kg crate is lifted 25.0 m upward by a force of N.How much work is being done by the applied force?W = FdW = (1000. N)(25.0 m)W = 2.50x104 J
11Practice ProblemsA satellite orbits the Earth in a circular orbit of radius = 10,000 km. How much work does the Earth do on the satellite?Surprisingly, none!The Earth does apply a force on the satellite, but it causes no displacement toward or away from the Earth.Also, neither the satellite’s kinetic nor its potential energy changes.
12Work-Energy Theorem A net force causes acceleration. If speed changes, KE changes.A net force acting over a distance does work.Sooo...Fnetd = KOr,W = K
13Practice ProblemsHow much work is done in accelerating a 500-kg vehicle from 0 m/s to 20 m/s?W = KKo = 0 JKf = ½ mv2 = 100,000 JWork done = 100,000 J
14Practice ProblemsWhat force is required to accelerate a 0.20-kg baseball from 0 m/s to 50 m/s over a distance of 1.25 m?W = KFd = Kf – KoFd = ½ mvf2 – ½ mvo2F(1.25 m) = 250 J – 0 JF(1.25 m) = 250 JF = 200 N (about 45 lbs of force)
15Practice ProblemsJust for fun (this is fun, right?), let’s solve the previous problem without using our knowledge of work or kinetic energy.Should we get the same answer?
16Practice ProblemsWhat force is required to accelerate a 0.20-kg baseball from 0 m/s to 50 m/s over a distance of 1.25 m?F = maDon’t know a, but we do know that:vo = 0 m/svf = 50 m/sx = 1.25 m
17Practice Problems vf2 = vo2 + 2ax F = ma Which method do you prefer? (50 m/s)2 = (0 m/s)2 + 2a(1.25 m)2500 m2/s2 = (2.50 m)aa = 1000 m/s2F = maF = (0.20 kg)(1000 m/s2)F = 200 NWhich method do you prefer?
18Practice ProblemsA force of 2600 N is applied at an angle 60º below the horizontal against a 200.-kg crate that is initially at rest on a frictionless surface.The crate moves 10.0 meters.What is the crate’s velocity when it has moved 10.0 meters?2600 N60º10.0 mvf = ???
19Practice ProblemsFirst, work out the horizontal component of the force.The only component that does work.Fhoriz = F*cosFhoriz = (2600 N)(cos60º)Fhoriz = 1300 N2600 N60º10.0 mvf = ???1300 N
20Practice Problems W = Fd = K (1300 N)(10.0 m) = 13,000 J K = 13,000 JKf = 13,000 J13,000 J = ½ mvf213,000 J = ½ (200 kg)vf2vf2 = 130 m2/s2vf = 11.4 m/s2600 N60º10.0 mvf = ???1300 N