OU PHYS 1205 - Work Energy and the CWE Theorem

Unformatted text preview:

Chapter 8Work, Energy, and the CWE TheoremScalar (“Dot”) Product Revisited A⋅B = (Axi + Ayj + Azk)⋅(Bxi + Byj + Bzk) = AxBx+AyBy+AzBz= AB cosθA·B is the magnitude of B in the direction of A multiplied by the magnitude of A, (or completely equivalently, the magnitude of A in the direction of B multiplied by the magnitude of B.)Define “money”CashSavingsCreditLandStocksBefore discussing the concept of energy it is insightful to talk about money.The total amount of “money” doesn’t change, but it can be transferred from one “system” to another.How much money a system has depends on the definition of the system.Is this a transfer of money from one system to another or not?Define “energy”It is hard to define energy. But we can say that the total amount of energy never changes. It is conserved. Energy can be transferred from one system to another.System• Motional energy = kinetic energy = K• Stored energy = potential energy = U• Internal energy (not discussed too much this semester)Mechanical energy = E = U + KWork = WEnergy transferEnvironmentTo understand this, we must develop the above concepts .Energy Transfer and Conservation(Initial energy of system) + (Energy transferred into system) = (Final energy of system) + (Energy transferred out of system)How would you write this if your system is the whole universe?(Initial energy of system) − (Final energy of system) = 0-or -(Initial energy of system) = (Final energy of system)-or -(Initial energy of system) − (Final energy of system) = (Net Energy transfer)Energy is conservedThe previous page can be expressed as the Work-Energy Theorem: ∆K = Wnet∆K = WC + WNC + Wext= −∆U −∆Eint+ WextEint= Ether + Echem+ Enucl+ Esound+ Elight+ …We will find next semester that to make this complete, we must add one more term, energy transferred by heat (Q)∆K = Q + Wnet∆K + ∆U + ∆Eint= Q + WextBut for this semester, we will set Eint= 0 and Q=0, so we will write the conservation of energy as∆K = Wnetor, more often:∆K + ∆U = WextIf energy is always conserved, what do we mean when we say that you should turn the lights off or drive a fuel efficient vehicle to “conserve” energy? An “isolated” system has no external forces acting on it.A “closed” system does not allow any energy or matter to be transferred into or out of the system.The universe is both a closed and isolated system.Definition of WorkWork is the product of the component of force along the direction of displacement times the magnitude of the distance.W = ∫F⋅drwhere W is the work done by the force F acting on an object, when the object moves a distance ds.If F is the net force acting on the object, then we have,Wnet= ∫Fnet⋅drThe SI unit of work is N⋅m which is given the name Joule.1 N⋅m = 1 JIf we write the integral in components, we getW = ∫F⋅dr = ∫Fxdx + ∫Fydy + ∫Fzdz(with F = Fxi + Fyj + Fzk, dr = dxi + dyj + dzk)If there is only motion in one dimension and the force constant, thenW= ∫Fxdx = Fx∆x = F cosθ∆xwhere θis the angle between the force, F, and the displacement, ∆x.If there is only motion in one dimension (x),W= ∫Fxdx∆K = WnetThe work-energy theorem only works when considering the net work done onan object!The definition of work gives the work done onan object by a force acting onan object. This will be the opposite sign to the work done the force. Be careful to differentiate between work done onan object and by an object, which usually means bya force.Wi= ∫Fi⋅dsProblem: (a) You pull a crate with a force of 98 N at an angle of 25° above the horizontal for a distance of 62 m. What is the total work done by you on the crate? (b) You now pull the crate with a force of 98 N at an angle of 25°above the horizontal for a distance of 62 m in the other direction. What is the total work you did on the crate in both directions?Problem: What work is done by a force F = (2.0x N)i + (3.0 N)j, with x in meters, that moves a particle from a position ri= (2.0 m)i + (3.0 m)j to a position rf= –(4.0 m)i – (3.0 m)j?You lift a barbell with a mass of 50 kg up a distance of 0.70 m. Then you let the barbell come back down to where you started. How much net work did you do on the barbell?A) − 340 JB) 0 JC) + 35 JD) + 340 JE) + 690 JInteractive QuestionYou lift a 10 N physics book up in the air a distance of 1 meter at a constant velocity of 0.5 m/s. The work done by gravity isA) +10 JB) − 10 JC) +5 JD) −5 JE) zeroInteractive QuestionHow much work is done by ……the team on the right if the rope doesn’t move?…the groom as he carries his bride over the threshold?…the person as he pushes the rock but it doesn’t move.Problem: A box weighing 23.0 kg slides down an inclined plane at a constant velocity as shown. The box slides 1.50 m. (a)What is the work done by the normal force, by gravity, and by friction? (b)What is the total work done on the box?37.0ºConsider the three different frictionless ramps shown below. If you push an identical box up each ramp at constant speed, rank the ramps in order from the one that would take the least work to the one that would take the most work. I II IIIA) I, II, III B) III, II, IC) II, I, III D) I, III, IIE) None of the aboveInteractive QuestionhSuppose you wanted to ride your mountain bike up a steep hill. Two paths lead from the base to the top, one twice as long as the other. Compared to the work you would do if you took the short path, the work you do along the longer path isA) four times as small.B) three times as small.C) half as small.D) the same.E) it depends on the time taken.Interactive QuestionSuppose you wanted to ride your mountain bike up a steep hill. Two paths lead from the base to the top, one twice as long as the other. Compared to the average force you would exert if you took the short path, the average force you exert along the longer path isA) four times as small.B) three times as small.C) half as small.D) the same.E) it depends on the time taken.Interactive QuestionAt a bowling alley, the ball feeder must exert a force to push a 5.0 kg bowling ball up a 1.0 m long ramp. The ramp raises the ball 0.5 m above the base of the ramp. Approximately how much force must be exerted on the bowling ball.A) 200 N B) 50 N C) 25 ND) 5.0 N E) Not enough info to tellF1.0 m0.5 mInteractive Question)(21)(21212022022000xxkkxkxxkxdxkxdFWxxxxxx−−=−−=′=′′−=′−=∫∫Variable ForcesConsider the force exerted


View Full Document

OU PHYS 1205 - Work Energy and the CWE Theorem

Download Work Energy and the CWE Theorem
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Work Energy and the CWE Theorem and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Work Energy and the CWE Theorem 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?