chapter 10 Flashcards
what are the two different systems they choose ?
System 1 has been chosen to consist of only the two particles; the forces are external forces.
Now consider the same two particles but with a different choice of system, system 2, where we’ve included the interaction within the system.
(no external forces)
what can you tell me about system 1 ?
System 1 is a restricted system of just the particles, so system 1 has only kinetic energy. All the interaction forces are external forces that do work. Thus system 1 obeys
ΔEsys=ΔKtot=WA+WB
what can you tell me about system 2 ?
System 2 includes the interaction, so system 2 has both kinetic and potential energy. But the choice of the system boundary means that no work is done by external forces. So for system 2,
ΔEsys=ΔKtot+ΔU=0
Notice that, for system 2, kinetic energy can be transformed into potential energy, or vice versa, but the total energy of the system does not change.
can you choose any system for calculations ?
The point to remember is that any choice of system is acceptable, but you can’t mix and match. You can define the system so that you have to calculate work, or you can define the system where you use potential energy, but using both work and potential energy is incorrect because it double counts the contribution of the interaction. Thus the most critical step in an energy analysis is to clearly define the system you’re working with.
what’s the symbol for gravitational potentail energy ?
UG
what’s the symbol for potential ?
U
Figure 10.2 shows a ball of mass m moving upward from an initial vertical position yi to a final vertical position yf. The earth exerts force F⃗ E on B on the ball and, by Newton’s third law, the ball exerts an equal-but-opposite force F⃗ E on B on the earth.
describe the 2 different system we can use
We could define the system to consist of only the ball, in which case the force of gravity is an external force that does work on the ball, changing its kinetic energy. We did exactly this in Chapter 9.
Now let’s define the system to be ball + earth. This brings the interaction inside the system, so (ignoring any gravitational forces from distant astronomical bodies) there’s no external work.
gravitational potential energy . equation ?
UG=mgy (gravitational potential energy)
Notice that gravitational potential energy is an energy of position. It depends on the object’s position but not on its speed. You should convince yourself that the units of mass times acceleration times position are joules, the unit of energy.
whats the equation we use for system 2 ?
UGf−UGi=mgyf−mgyi
Example 10.1 Launching a pebble
Rafael uses a slingshot to shoot a 25 g pebble straight up at 17 m/s. How high does the pebble go?
15m
whats the equation we use for system 2 ?
UGf−UGi=mgyf−mgyi
does it matter where you put your point of reference ?
no. as long you are consistent throughout the problem
mechanical energy ?
def: the total macroscopic kinetic and potential energy
is telling us that—in this situation—the mechanical energy does not change as the object undergoes vertical motion. Whatever initial mechanical energy the system had before the vertical motion, it has exactly the same mechanical energy after the motion. Kinetic energy may be transformed into potential energy during the motion, or vice versa, but their sum remains unchanged.
first statement of law of conservation of energy ?
Ki+UGi=Kf+UGf
Now this is a highly restricted situation: only the gravitational force, no other forces, and only vertical motion.
For example, Figure 10.8 shows an object sliding down a curved, frictionless surface
what the change in y
and tell me about the normal force in this situation and it affect the energy ?
he change in gravitational potential energy of the object + earth system depends only on Δy, the distance the object descends, not on the shape of the curve. But now there’s an additional force—the normal force of the surface. Does this force affect the system’s energy? No! The normal force is always perpendicular to the box’s instantaneous displacement, and you learned in Chapter 9 that forces perpendicular to the displacement do no work. Forces always perpendicular to the motion do not affect the system’s energy. They can be ignored during an energy analysis.
