Announcements
• William Henry Harrison (234)
9th President
March 4, 1841 - April 4, 1841
Applied Physics
• The Conservation of Energy
Quiz is available now!
• The Power/Efficiency Quiz went up this morning at
02-09-07
Power and Efficiency
8 AM.
• Wear comfortable shoes to lab on Thursday. We
will be climbing stairs to collect data.
• Homework Problems: 4.22, 4.24, 4.27, 4.30
Conservation of Energy
• There is one equation to remember.
• KE + PE + W + OE = KE + PE
i
i
f
f
Another Example Problem
A skier coasts down a very smooth, 10.0-m high slope. If
the speed of the skier at the top of the slope is 5.0 m/s, what
is his speed at the bottom of the slope?
KEi + PEi+ W + OE = KEf + PEf
0.5mvi2 + mgh+ 0 + 0 = 0.5mvf2 + 0
0.5mvi2 + mgh = 0.5mvf2
0.5m 0.5m 0.5m
(vi2
vi2 + 2gh = vf2
+ 2gh)1/2 = vf
[(5.0 m/s)2 + 2(9.8m/s2)(10.0 m)]1/2 = vf
14.9 m/s = vf
Power
Power Example
A girl consumes 2000 Calories (8.4 x 106 J) of energy per
day while maintaining a constant weight. What is the
average power she produces in a day?
• Power is the rate of energy use.
• Power = Energy / time
• Units: J/s = Watts (W)
Efficiency
!=
!=
Useful Energy
x 100%
Energy Used
(Useful Energy)/time
Useful Power
x 100% =
x 100%
Power Used
(Energy Used)/time
Energy
8.4x106 J
Power = time = (24 hr)(60 min)(60 s) = 97.2 W
Example
A 70 kg hiker climbs a mountain 1000 m high in 3 hours.
(a) Calculate the external (useful) work done by the hiker.
(b) Assuming that the work is done at a steady rate during
the 3-hour period, calculate the hiker’s average power
expenditure during the climb.
(c) If the hiker burns 1600 kcal of food energy during the
climb, what is efficiency of the hiker’s body?
Example
(a) Consider the hiker at the bottom of the hill and the top
of the hill:
KEi + PEi + W + OE = KEf + PEf
0 + 0 + W + 0 = 0 + mgh
W = mgh = (70 kg)(9.8 m/s2)(1000 m) = 686,000 J = 686 kJ
(b) Power = Energy/time
The time will be (3 hr)(60 min/hr)(60 s/min)= 10,800 s
Power = (686,000 J)/(10,800 s) = 64 W
(c) The “Useful Energy” is the PE gained.
The “Energy Used” is the food energy burned.
Energy Used = (1600 kcal)x(4186 J/kcal) = 6,697,600 J
! = [(686,000 J)/(6,697,600 J)] x 100% = 10.2%