Review of the Concept of Work

At this point it is worth mentioning once again that work derives from Newton's definition of force in his Second Law. As such, we must, strictly speaking, confine our definition of work to particles in inertial frames. Finite 3-dimensional objects are more complicated, but this definition works pretty well as an approximation for many objects. Soon we will be able to replace the approximation for three-dimensional objects by an exact statement about the motion of a particular point for an object, a point we will call the center-of-mass. More on that later.

• Example 1:

  We might consider an example to show how this definition works. Suppose that you pick up a box from the floor, hold it in your arms, and carry it across the room before setting it back on the floor. What is the net work done by you in this whole operation?

  Solution: First, in picking up the box, you do net work W_up = m_boxgh where h is the height at which you hold the box in your arms. As you walk across the floor, you do no work as you are not applying any horizontal forces to the box (assuming you walk without acceleration) and the only other forces on the box are in the vertical direction and therefore perpendicular to the box's displacement. At the other side of the room you place the box down. If you are not dropping the box, but let it down at constant velocity, then you must apply a force of magnitude m_boxg to the box over the distance -h, so the work done going down is W_down = -m_boxgh. The net work for the whole operation is W_net = W_up + W_down = 0.

  The table below summarizes the work done by you and by the Earth's gravity for this operation.

  Work Done
  By you By gravity pick up: +mgh -mgh walking: 0 0 put down: -mgh mgh

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  Lecture Question 1

  The figure below shows four situations in which a force acts on a box while the box slides to the right a distance d. The floor is frictionless. The magnitudes of the forces are identical and their directions are as shown by the arrows. Rank the situations according to the work done on the box by the force shown for that box. The ranking should be from most positive to most negative.
  Choose the correct order from below.
  1. (a.) then (b.) then (c.) then (d.)
  2. (d.) then (c.) then (b.) then (a.)
  3. (c.) then (d.) then (b.) then (a.)
  4. (b.) then (a.) then (d.) then (c.)
  5. (d.) then (b.) then (c.) then (a.)

Potential Energy and Total Mechanical Energy

As noted in the textbook, the work done by gravity is distinct from work done by friction. Namely, if we lift, with constant velocity, an object in a gravitational field, and then lower it to its original position, the total work done is zero since work depends only on the change in vertical height. We call such forces conservative. Friction is not conservative since if we drag a block on any surface with friction, the work done depends only on how far we dragged it, not where the path of travel began and ended. Conservative forces have the nice property of being able to be defined in terms of a potential energy. The usual definition of potential energy is through the work-energy theorem as for kinetic energy, i.e. W = U_i - U_f. We can define a change in potential energy also through the action of a force over a distance. The potential energy is useful in that it keeps us from calculating the work done by a force when the path is complicated, or tedious to follow. Since the work done depends only on the initial and final positions, the change in potential (and kinetic) energy depends only on change in position as well. Through the work energy theorem, we have K_f - K_i = W = U_i - U_f ==> K_i + U_i = K_f + U_f = constant = E where E is called the total mechanical energy.

In general we can define any combination of conservative and non-conservative forces acting on particles as obeying the principal of conservation of mechanical energy. This general rule is expressed mathematically as

The expression states that any non-conservation of energy is due to work done by forces which do not conserve mechanical energy. Generally, this energy shows up as other, harder-to-measure, forms of energy such as heat. The principle of total energy conservation when all forms of energy are considered took many years to prove and required input from other fields of science, principally chemistry.

• Example 2:

  Why are roller coaster loops made into egg-shaped ovals rather than circles? First, start with the fact that the ride begins by lifting the cars to a large height, then allowing gravity to power the rest of the ride. We can use conservation of energy and and knowledge of circular motion to predict what would happen for a circular loop as depicted below.

  

  If we follow this prescription, then, setting the origin of our coordinate system at ground height, the initial mechanical energy of the coaster car (assume mass m) is

  K₀ + U₀ = 0 + mgh

  and the mechanical energy, neglecting friction, at the bottom of the coaster loop is

  K_bot + U_bot = ½ mv_bot² + 0

  At the top of the coaster loop we have

  K_top + U_top = ½ mv_top² + mg(2R)
  1. What is the minimum height of the coaster for passengers to be assured of completing the loop?

     Solution:
     The normal force must be just greater than zero at all points around the loop else it means the car loses contact with the track or the passenger with the seat. The force diagrams are shown in the figure above. We want, at the top,

     -FN, min. - mg = mvtop2 R 0 - mg = mvtop2 R Þ vtop =   __ ÖRg

     This sets the absolute minimum speed for the coaster car. To get to this minimum speed we must have h greater than a minimum height set by energy conservation.

     K0 + U0 = Ktop + Utop 0 + mghmin = 1 2 mvtop2 + mg(2R) Þ ghmin = 1 2 Rg + 2Rg Þ hmin = 5 2 R

  2. What ``g'' forces act on the passengers at the bottom of the loop given this minimum height?

     Solution:
     To get that, we first need the speed at the bottom assuming the minimum height for h.

     K0 + U0 = Kbot + Ubot 0 + mghmin = 1 2 mvbot2 + 0 Þ 5 2 gR = 1 2 vbot2 Þ vbot =   ___ Ö5gR

     The normal force at the bottom of the loop is determined by the necessity of satisfying the centripetal condition (i.e. for the car to stay on the track, the gravitational and normal forces must add up to a net force which satisfies the centripetal condition)

     FN - mg = mvbot2 R Þ FN = mg + m(5gR) R = 6mg

     So the poor passenger at the bottom experiences 6 g's. Remember that this is for the minimum height. Many people would quickly black out at this acceleration.

Lecture Question 2

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