Physics 218 Alexei Safonov Lecture 10: From Work to Energy.

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Presentation transcript:

Physics 218 Alexei Safonov Lecture 10: From Work to Energy

Other tricks What is minimum velocity for the car not to fall down? –Radius, mass are given –Key question: what tells you that the car may not make it through the top point? Normal force! How about a problem with a ball on a string that you rotate in vertical plane with constant speed and you want it to swing without sagging at the top? – What force should you look at instead? What if it is a bucket with a rock in it and you don’t want it to fall on you? –What force?

Conical Pendulum A small ball of mass m is suspended by a cord of length L and revolves in a circle with a radius given by r = Lsin . 1.What is the velocity of the ball? 2.Tension F T ? 3.Calculate the period of the ball.

Overview: Chapters 6 & 7 Combine Chapter 6 & 7 into three and one third lectures Today we’ll cover Work: Intuitive understanding The math and multiple ways to calculate work

Work The word “Work” means something specific in Physics (~ like Force) The amount of Work we do is the amount of Forcing we do over some distance Example: If we are accelerating a car for 1 mile, then there is a force and a distance –We do Work

Calculating the work Work is done only if the force (or some component of it) is in the same (or opposite) direction as the displacement Work is the force done Parallel to the displacement –Think of it as “amount of displacement should be affected by the force if it is doing any work” Pushing a car to speed it up or to slow it down

Work for Constant Forces The Math: Work can be complicated. Start with a simple case. For constant forces, the work is …(more on this later) W=F. d

1 Dimension Example Pull a box with a constant force of 30N for 50m where the force and the displacement are in the same direction How much work is done on the box? W = F. d = 30N. 50m= 1500 N. m = 1500 Joules

SP Checkpoint A box sits on the horizontal bed of a truck accelerating to the left as shown. Static friction between the box and the truck keeps the box from sliding around as the truck drives. The work done on the box by the static frictional force as the accelerating truck moves a distance D to the left is? A.Positive B.Zero C.Negative D.Depends on the speed of the truck.

SP Checkpoint A box sits on the horizontal bed of a truck accelerating to the left as shown. Static friction between the box and the truck keeps the box from sliding around as the truck drives. The work done on the box by the static frictional force as the accelerating truck moves a distance D to the left is? A.Positive B.Zero C.Negative D.Depends on the speed of the truck.

Work done and Work experienced Something subtle: The amount of work YOU do on a body may not be the same as the work done ON a body Only the NET force on the object is used in the total work calculation –Add up all forces to get net force, calculate work done on the object … Or add up the work done on an object by each force to find the total work done! –But be careful with signs!

Examples While you are lifting up a bottle with mass m, the bottle moves a distance d with constant velocity. As you lift it: –What is the force you exert? –What is the work done by you? –What is the work done by gravity? –What is the net work? You push a box with Force F on a rough floor with coefficient of friction  for a distance d, and the box moves with constant velocity. As it moves: –What is the work done by you? –What is the work done by friction? –What is the net work?

You take a box of mass m and move it (with ~zero constant velocity) to height h and then lower it back (also with ~zero velocity). As it moves, we calculate the work done on it. The total work done on the object is … A.mgh at the top point, 2mgh when returns to the lower point again B.mgh at the top, but zero when at lower point again C.Zero at the top and zero at the lower point again D.2mgh at the top, 4mgh by the time it returns to lower point

SP Checkpoint Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of of a string. In which case does the object have the biggest total work done on it by all forces during its motion? A.Free Fall B.Incline C.String D.Same

SP Checkpoint Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of of a string. In which case does the object have the biggest total work done on it by all forces during its motion? A.Free Fall B.Incline C.String D.Same

Force Displacement 2 Dim: Force Parallel to Displacement Force Displacement Rotate F || = Fcos  W = F || d = F. d = Fdcos  where  is the angle between the net Force and the net displacement. You can think of this as the force component in the direction of the displacement.

Examples Holding a bag of groceries in place –Is it heavy? –Will you get tired holding it? –Are you doing “Work?” Moving a bag of groceries with constant speed across a room –Is it heavy? –Will you get tired doing it? –Are you doing “Work?” Lifting a bag of groceries a height h with constant speed? –Work by you? –Work on the bag?

Groceries: With the math Holding a bag of groceries –Work by you on bag: W=Fdcos  =(mg)*(0)*cos  = 0 –Work on bag W=Fdcos  =(0)*(0)*cos  = 0 Moving a bag of groceries with constant speed across a room –Force exerted by you = mg, Net Force on bag = 0 –Work on bag= F. d = Fdcos  =0*dcos  =0 –Work exerted by you =Fdcos  =mgd*cos(90 0 )=0 Lifting a bag of groceries a height h with constant speed? –Work on bag = Fd*cos  = (0)*h*(0 0 ) = 0 –Work by you =Fdcos  =(mg)hcos(0 0 )=mgh

Work in Two Dimensions You pull a crate of mass M a distance X along a horizontal floor with a constant force. Your pull has magnitude F P, and acts at an angle of . The floor is rough and has coefficient of friction . Determine: The work done by each force The net work on the crate  X