1018
Wnc - `dKE - `dPE = 0
Wnc is work by net nonconservative force on the system.
If `dPE is 0 then the only forces are nonconservative, and this is equivalent to
`dWnet = `dKE,
which is the work-energy theorem.
`dPE is the work done by the system against the net nonconservative force, which is the negative of the work done by the net nonconservative force acting on the system.
`dWnetOn = `dWnonconsOn + `dWconsOn.
`dWnonconsOn is expressed in Giancoli as Wnc.
`dWconsOn = -`dPE (see above).
`dWnetOn is thereforeexpressed as `dWnc - `dPE, where `dWnc might be expressed as Wnc and means `dWnonconservativeOn.
`dWnetOn = `dWnc - `dPE = `dKE so
`q For an object coasting up or down a hill in the presence of
friction, gravity and the
normal force:
`l When an object slides either up or down a hill the force of friction is in
the direction
opposite its motion. The frictional force acting on the object therefore does
negative work on it. This force is nonconservative, so Wnc, the work done by the
nonconservative force acting on the object, is negative.
The normal force exerted by the hill on the object is perpendicular to the direction of motion, so there is no displacement in the direction of the normal force. The normal force therefore does no work on the object.
The gravitational force is conservative, and it has a
component down the hill. Therefore if the object coasts down the hill the
gravitational force does positive work on it. In this case the object is losing
gravitational PE (gravity is doing work on the object so gravitational PE is
being expended). If it coasts up the hill the gravitational force does negative
work
on the object, and the object gains gravitational PE.
`q For an ideal rubber band projecting a sliding object up an incline:
`y Part of the KE attained by the object at the end of the snapback is
dissipated against friction as the object slides up the incline.
Wnc is negative (friction working against motion), `dKE is negative (the thing comes to rest) and `dPE is positive (PE increases with altitude).
`l When an ideal rubber band is first pulled back then released to project a sliding object up an incline, positive work must be done on the rubber band by a force acting in the direction of the stretch. This does positive work on the rubber band, which an ideal rubber band will entirely store as elastic potential energy. When the rubber band is released some of this potential energy is lost to friction while the rubber band snaps back, and since the object slides a ways up the hill as the rubber band snaps back some of its elastic PE is also converted to increased gravitational PE. The remaining energy is converted to the KE of the object. Part of this KE is dissipated against friction as the object continues to slide up the hill, and the remainder is converted to another increase in gravitational PE.
*&*&
`q For a thin trapezoid of a force vs. position graph, which
of the following are true?
`l The area under a thin trapezoid of a force vs. position
graph is equal to the average
altitude of the trapezoid multiplied by its width. Altitudes of the trapezoid
represent
forces, so the average altitude will represent the approximate average force. If
the
trapezoid is thin the slope at the top of the trapezoid won’t change much so the
segment at
the top of the trapezoid will be nearly linear, and the average altitude will be
close to
the actual average force. {}{}The width of the trapezoid represents change in
position or
displacement. So the area of the trapezoid represents the approximate product of
average
force with displacement, which is the approximate work done over that
interval.{}{}The slope
of the trapezoid is rise / run = change in force / change in position = average
rate of
change of force with respect to position.