Alphabetical Order. A dictionary is a storehouse of information. It gives a list of the words of language in alphabetical order and explains what they mean. ... Tongue Twisters. During your holidays, learn these tongue twisters and have fun. You may
Alphabetical Order. A dictionary is a storehouse of information. It gives a list of the words of language in alphabetical order and explains what they mean. ... Tongue Twisters. During your holidays, learn these tongue twisters and have fun. You may
There are only two mistakes one can make along the road to truth; not going all the way, and not starting.
Idea Transcript
Work and Friction
Force Parallel to Motion Only the parallel force does work. W=Fd Example: A force of 30 N is applied to a box over a distance of 15 meters. What is the work done? Answer: W=FD; 450 J
Force Not Parallel Ө
If Ө = 30° and the Force is 20 N. How far would the box move if 100J of work were expended? Answer 5.77 meters.
Force going down Ө
What is the same force is pointing down? How far will the block move? (We are neglecting friction) Answer: 5.77 meters
Adding in Friction F Ө
F
mg
Box weighs 20 N. θ = 30° F=3N D = 50 m If the box is moving at a constant velocity, how much work does the Force do? How much work does friction do?
How much work does the Force Do? Remember, only the parallel forces do work. So… W = Fd W = 50m ( 3N ) cos 30° W = 130 J
How much work does Friction do? First, find Friction: F = µN = F cosθ So W = Fd W = 50 ( 3N ) cos 30° W = 193.19 J
Pushing Force
F Ө
mg
Box weighs 20 N. θ = 30° F = 3N D = 50 m Now, how much work does the Force do?
Solution Once again, the velocity does not matter. W = Fd Fx = F cosθ d = 50m W = 50m (3N) cos θ W = 193.19 J
Friction Example A 25 kg crate of chocolate is sitting on a loading. It needs to be pulled 10 meters to the store. The coefficient of kinetic friction between the box and the sidewalk is .22. How much work will a man exert if he pulls the box with a 60N force at an angle of 30° How much work will friction exert? What, if any, will be the acceleration of the box?
Solution N Ө
F
mg
We Start Solving Since only the horizontal force contributes to work: Fx = F cosӨ W = Fd W = Fd cos θ W = 10m (60) cos 30° W = 519.6 J
Friction’s Work Find the normal force: N = mg – F sinӨ F = μN F = μ(mg - FsinӨ) W = D *μ(mg - FsinӨ) W = 473 J
What is the acceleration? Force going forward: F cos θ Friction: µ (mg – F sinθ) Calculate each of these and find the difference F cosθ = 60 cos 30° F = 52.0N Friction: .22 ( 25*9.8 – 60 sin 30° ) Friction = 47.3N F = ma 4.7 = 25 a a = .188 m/s2
Bonus Question How fast is the box moving when it gets to the store? We get 1.93 m/s.