# Work.

## Presentation on theme: "Work."— Presentation transcript:

Work

Work When a force is done on an object, energy is transferred to the object. The amount of energy transferred is called work.

Work If you apply a 200.-Newton force to move a box 50.0 meters, how much work have you done?

Work Symbol: W SI Unit: Joule Equation: W = Fd
Force must be in same direction as movement. Work only happens if a force causes movement.

Is 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.

Is 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.

Is 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.

Is it Work? You push a lawnmower across the lawn.
Yes! Only the horizontal component of your applied force does work on the lawnmower, however.

Work 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!

Practice Problems A 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 = Fd W = (1000. N)(25.0 m) W = 2.50x104 J

Practice Problems A 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.

Work-Energy Theorem A net force causes acceleration.
If speed changes, KE changes. A net force acting over a distance does work. Sooo... Fnetd = K Or, W = K

Practice Problems How much work is done in accelerating a 500-kg vehicle from 0 m/s to 20 m/s? W = K Ko = 0 J Kf = ½ mv2 = 100,000 J Work done = 100,000 J

Practice Problems What 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 = K Fd = Kf – Ko Fd = ½ mvf2 – ½ mvo2 F(1.25 m) = 250 J – 0 J F(1.25 m) = 250 J F = 200 N (about 45 lbs of force)

Practice Problems Just 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?

Practice Problems What 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 = ma Don’t know a, but we do know that: vo = 0 m/s vf = 50 m/s x = 1.25 m

Practice Problems vf2 = vo2 + 2ax 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)a a = 1000 m/s2 F = ma F = (0.20 kg)(1000 m/s2) F = 200 N Which method do you prefer?

Practice Problems A 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 N 60º 10.0 m vf = ???

Practice Problems First, work out the horizontal component of the force. The only component that does work. Fhoriz = F*cos Fhoriz = (2600 N)(cos60º) Fhoriz = 1300 N 2600 N 60º 10.0 m vf = ??? 1300 N

Practice Problems W = Fd = K (1300 N)(10.0 m) = 13,000 J
K = 13,000 J Kf = 13,000 J 13,000 J = ½ mvf2 13,000 J = ½ (200 kg)vf2 vf2 = 130 m2/s2 vf = 11.4 m/s 2600 N 60º 10.0 m vf = ??? 1300 N