Transcript Lecture 10b

Sect. 8-7: Gravitational Potential Energy &
Escape Velocity
As we’ve already seen, far from the surface of the
Earth, the force of gravity is not constant:
From the formal definition of work
we had earlier, the work done on an
object moving in the Earth’s
gravitational field is given by:
Integrating gives:
Because the value of the integral depends only on the
end points, the Gravitational Force is Conservative &
we can define a gravitational potential energy:
Ex. 8-12: Package Dropped from a High Speed Rocket
A box of empty film canisters is allowed to
fall from a rocket traveling outward from
Earth at a speed of 1800 m/s when 1600 km
above the Earth’s surface. The package
eventually falls to the Earth. Calculate its
speed just before impact. (Ignore air resistance.)
If an object’s initial kinetic energy is equal to
the potential energy at the Earth’s surface, its
total energy will be zero. The velocity at which
this is true is called the ESCAPE VELOCITY.
For Earth this is:
Example 8-13: Escaping the Earth or the Moon
a. Compare the escape velocities of a rocket from the
Earth and from the Moon
b. Compare the energies required to launch the rockets.
Moon: MM = 7.35  1022 kg, rM = 1.74  106 m
Earth: ME = 5.98  1024 kg, rE = 6.38  106 m
Sect. 8-8: Power
• Power  The rate at which work is done or the rate
at which energy is transformed.
• Instantaneous power:
• Average Power:
P = (Work)/(Time) = (Energy)/(time)
SI units: Joule/Second = Watt (W) 1 W = 1J/s
British units: Horsepower (hp). 1hp = 746 W
“Kilowatt-Hours” (from your power bill). Energy!
1 KWh = (103 Watt)  (3600 s) = 3.6  106 W s = 3.6  106 J
Example 8-14: Stair Climbing Power
A 60-kg jogger runs up a flight
of stairs in 4.0 s. The vertical
height of the stairs is 4.5 m.
.
a. Estimate the jogger’s power
output in watts and horsepower.
b. How much energy did this
require?
Power is also needed for acceleration & for
moving against the force of friction.
The power can be written in terms of the net
force & the velocity:
• Average Power: P = W/t
• Its often convenient to write power in
terms of force & speed. For example,
for a force F & displacement d in the
same direction, we know:
W=Fd
 P = F (d/t) = F v = average power
v  Average speed of object
Example: Power Delivered by an Elevator Motor
An elevator car of mass me = 1,600 kg,
carries people, mass mp = 200 kg. A
constant friction force fk = 4,000 N acts
against the motion.
(A) Find the power needed for a motor
to lift the car + passengers at a
constant velocity v = 3 m/s.
(B) Find power needed for a motor to
lift the car at instantaneous velocity
v with an upward acceleration
a = 1.0 m/s2.
Example 8-15: Power needs of a car
Calculate the power required of a 1400-kg car to do the following:
a. Climb a 10° hill (steep!) at a steady 80 km/h
b. Accelerate on a level road from 90 to 110 km/h in 6.0 s
Assume that the average retarding force on the car is FR = 700 N.
a.
∑Fx = 0
F – FR – mgsinθ = 0
F = FR + mgsinθ
P = Fv
b.
Now, θ = 0
∑Fx = ma
F – FR= ma
v = v0 + at
P = Fv