WORK ENRGY AND POWER

Work - Giancoli, Ch 6, 1-3 visit the Physics Classroom for this topic
1. Define work.
· In physics, work is defined as a force acting upon an object to cause a displacement.
· In general terms, the amount of work performed is calculated as the product of the force and the displacement.
· To see how the product is calculated, consider the example of a rope pulling at an angle, q, above the floor on a box. The result is that the force of the rope, F, moves the box horizontally a distance d.

When looking at the force, F, that acts, only a portion of the force is responsible for the displacement, d. The vertical component of the force (F sin q) has no effect on the motion. It is only the horizontal component, F cos q, that moves the box. So, to find the amount of work done by F, we multiply the horizontal component of F times the magnitude of the displacement, d.

W = Fd cos q

· In words, to find the work done by a force to produce a displacement, multiply the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them.
· The unit of work is the Nm or Joule.
2. Determine the work done by a non-constant force by interpreting a force–displacement graph.
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Suppose a constant force acting horizontally moves a box across the floor. If we were to make a graph that shows the magnitude of the pulling force as a function of the position of the box, it would look like the graph below.







We know that in this case, the work done by the force is Fd. Looking at the graph, we see that the area of the shaded rectangle is also Fd. This suggests that the work done by a force to produce a displacement, d, is the area of a force-displacement graph.
· Now suppose that the force that is doing the work does not remain constant, but varies with the position of the object, like the second graph above. The work done in this case is still the area of a force-displacement graph, as shown by the shaded region of the graph.
3. Solve problems involving the work done on a body by a force.

· See Giancoli, Ch 6 problems #5, 7a, 9, 13
· PLUS, do the following problem: The force required to stretch an ideal spring varies with the "stretch" of the spring as F = kx, where k is called the spring constant and x represents the amount of stretch from the equilibrium position of the spring. Show that the work necessary to stretch a spring from a position, x1, to a new position, x2, is given by the expression W = 1/2 kx22 - 1/2 kx12. Hint - look at a graph of the force as a function of position of the spring.

Energy and power - Giancoli, Ch 6, 3-10 visit the Physics Classroom for this topic
4. Define kinetic energy.
· Suppose the result of a constant force acting to move an object from rest through a distance, d, is that the object acquires a final speed, v. We know that the work done by the force is given by W = Fd = mad. We also know from kinematics, that the relation v2 = u2 + 2ad = 2ad (because it starts from rest) can be used to describe this motion. If we solve the kinematic relation for the quantity ad = v2/2 and substitute into the expression for the work done, we get W = Fd = mad = mv2/2.
· The quantity, , is called the Kinetic Energy (K) of the object. This is the form of energy related to motion.
· The result of doing work on the object is to change its energy. In this case it gained kinetic energy (= ).



5. Describe the concepts of gravitational potential energy and elastic potential energy.
· Suppose a box of mass m at rest on the ground is lifted to a height h above the ground so that its final speed is zero. How much work is required to lift the box? The force that we have to overcome to lift the box is its weight, mg. The work to lift it is then found using the definition of work, W = Fd = mgh.
· The quantity, mgh, is called the gravitational potential energy (Ug) of the box. This is the form of energy related to a change in position of the box. It has this energy because we moved it, but it is not in motion.
· The force required to stretch a spring varies with the position of the spring as F = kx. This is called Hooke's Law. You showed in the homework problems for work above that the work done by this force to stretch the spring a distance x is given by the expression W = area under graph = .
· The quantity, , is called the elastic potential energy (Uel­) of the spring. This is also a form of energy acquired by the spring because of the work done to change its position.
6. State the principle of conservation of energy.
· The quantity U + K (potential + kinetic energy) of an object is called the mechanical energy (E) of the object.
· If no external forces do work on the object, then the mechanical energy (E) of the object remains constant. This is called conservation of mechanical energy.
· If external forces are present (such as friction or some outside push or pull), then the work done by the external force will equal the change in the mechanical energy of the object. W = DE. This is called the work-energy theorem and describes how energy is given to or taken away from objects. Since work results in a change of energy, energy must have the same unit of measure, Joules, as work.
7. List different forms of energy and describe examples of the transformation of energy from one form into another.
· Energy exists in many forms (heat, nuclear, solar, etc.) which can be ultimately reduced to the kinetic and potential energies of constituent particles in objects.
· Conservation of energy states that in the absence of external forces the mechanical energy of an object remains constant. This means that DE = DU + DK = 0. In other words, DU = - DK. If we allow a ball to roll down a hill influenced only by the force of gravity then the potential energy will decrease as the ball moves down the hill. According to energy conservation, this means that the kinetic energy of the ball changes in the opposite way (increases) by the same amount. The potential energy of the ball has been transformed into kinetic energy.
8. Define power.
· When a quantity of work, W, is performed in a time, Dt, the ratio W / Dt is called the power developed.
· Since we have shown that W = DE, we can also define the quantity DE / Dt to be power.
· Power is the rate at which work is done or energy is changed. The unit of power is the Watt (W) = Js-1.
· A special case is when a constant force acts on object. P = W / Dt = (F Ds) / Dt = Fv.
9. Define and apply the concept of efficiency.
· In a perfect world, forces like friction are assumed to be negligible and the concept of efficiency is not necessary. However, in the real world, friction exists. Suppose that a box is pushed up a ramp. The box gains potential energy = mgh. However, because of friction, we have to push with more force than the weight of the box to get it up the ramp. The useful work is the change in potential energy of the box. The actual work is the amount done to slide it up the ramp. In general, the actual work is greater than the useful work.
· The efficiency is defined as the ratio of useful work / actual work.
10. Solve work, energy and power problems.

· See Giancoli, Ch 6 # 19, 23, 29, 35, 40, 41, 63, 74

Projectile Motion - Giancoli, Ch 3, 5-7 visit the Physics Classroom for this topic
1. State the independence of the vertical and horizontal components of motion for a projectile in a uniform field.
· Vertical forces do not affect horizontal motion. This means the horizontal component of projectile motion is motion with constant speed because there are no horizontal forces acting.
· The vertical component of projectile motion is accelerated motion (affected by the acceleration of gravity). At every point in the trajectory of a projectile, the acceleration vector is vertical (downward) and has the value 9.8ms-2.
· The key to analyzing projectile motion is to consider the two perpendicular components separately.




2. Describe the trajectory of projectile motion as parabolic in the absence of friction.
· The combination of accelerated vertical motion and constant horizontal motion produces a trajectory through the air in the shape of a parabola. (This can be proved mathematically, but is not required here.)
· The key to this process is that both motions are linked through time. It takes the same amount of time to complete both motions.
3. Solve problems on projectile motion. Problems may involve projectiles launched horizontally or at any angle above or below horizontal. Applying conservation of energy may provide a simpler solution to some problems than using projectile motion kinematics equations.
· The key to problem solving is to write the equations of motion for the horizontal and vertical components of the motion separately and remember that they are linked through time.
· Conservation of Energy can used to relate speed and height at any two points in the trajectory of an object.

· See Giancoli, Ch 3 # 37, 39, 43, 51