Note: Include
units in all problems.
1. A
woman weighing 100
pounds walks up 3 flights of stairs, each flight having 10 steps. Each
step is
6.0 inches high. What is her increase in gravitational potential
energy?
Where does this energy come from?
2. A man
having a mass of 80 kg is
running at 5.0 m/s. What is his kinetic energy?
Where does
this energy come from?
3. How much
heat will it take to raise the
temperature of 10.0 kg of water 1.0 Centigrade degree?
Express in
terms of kilocalories, calories, and joules. NOTE: 1 kilocalorie =
4,186
joules.
4. How much
heat will it take to raise the
temperature of 20 pounds of water 1.0 Fahrenheit degree?
Express
in terms of BTU's and ft lbs. NOTE: 1
BTU = 778 ft lb
5. How much
does it cost to operate a 125
W sleeping blanket for an 8.0-hour night's sleep? (Assume it’s
on all
the time.) The cost is 9.0 cents per kilowatt- hour.
6. How much
will a loaf of bread, costing
$1.00 per loaf in 1982, cost in the
year 2002
if the rate of inflation continued at 14% per year? Note: This
is just
20 years from that date!
Physics 1110 Problem
Set 2 Solution Sheet
1. Weight = 100lbs
Height = (3flights)(10
steps/flight)(0.50 ft/step) = 15.0 ft.
Increase in
GPE = weight x height = (100lbs)(15.0
ft) = 1500ft lb The
increase
in gravitational potential energy comes from work done by the woman
(which in
turn comes from the food she eats which in turn comes from the sun
which grows
the food).
2. KE = ½ mv2
= ½ (80 kg)(5.0 m/s)2
= ½ (80)(25) kg
m2/s2
Or
KE =1000 joules
The energy
again comes from energy (from
food) stored in the muscles of the man.
3. Heat
required = (10 kg)(1.000
kilocalorie/kg Cº)(1.0 Cº) = 10
kilocalories
10
kilocalories =10,000 calories = 10
Calories (namely 10 Food Calories)
Also 10
kilocal=
(10 kilocal/1) (4,186j/kilocal) = 41, 860 joules = 42,000 joules
(using
Significant Figures)
4. Heat
required = (20lbs)(1.000
BTU/lb Fº) (1.0 Fº)= 20 BTU
(just barely 2 sig. figs.)
Also 20
BTU = (20 BTU/1) (778ft
lb/BTU) = 15,560 ft lb = 16,000 BTU (keeping 2 sig. figs.)
5. P= 125
W = 0.125 KW T= 8.0
H W= Pt=(0.125
KW)(8.0 H) = 1.00 KWH
Cost = (1.00
KWH)(9.0 cents/KWH)
= 9.0 cents
6. Doubling
Time =70/P = 70/14 = 5yrs
1982 $1.0 1987 $2.0 1992 $4.0
1997 $8.0
2002
$16.0