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Work

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

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

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**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.

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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.

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**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.

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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.

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**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.

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**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!

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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

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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.

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**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

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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

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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)

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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?

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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

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**Practice 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)a a = 1000 m/s2 F = ma F = (0.20 kg)(1000 m/s2) F = 200 N Which method do you prefer?

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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 = ???

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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

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**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

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Work, distance and force

Work, distance and force

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