Note: Include
units in all problems.

1**.
**A woman weighing 10__0__ 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 8__0__ 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 2__0__ 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 12__5__ 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 $1.0__0__ per loaf in 1982, cost in the year 2002 if the rate of inflation
continued at
1__4__% per year? Note: This is just 20 years from __that date! __

Physics
1110 Problem
Set 2 Solution Sheet

1. Weight = 10__0__lbs

Height
= (3flights)(10 steps/flight)(0.5__0__
ft/step) =
15.__0__ ft.

2. Increase in GPE = weight x height = (10__0__lbs)(15.__0__ ft) = 15__0__0ft 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).

3.
KE = ½ mv^{2 }= ½ (8__0__ kg)(5.__0__
m/s)^{2 }= ½ (8__0__)(2__5__) kg m^{2}/s^{2 }Or KE =10__0__0
joules

The
energy again comes from energy (from food) stored in the muscles of the
man.

4. Heat required = (10 kg)(1.00__0__
kilocalorie/kg Cº)(1.__0__ Cº) = 1__0__ kilocalories

10
kilocalories =10,000 calories = 1__0__
Calories (Food
Calorie)

Also
1__0__ kilocal= (10 kilocal/1)
(4,186j/kilocal) =
41, 860 joules = 4__2__,000 joules (using Significant Figures)

5. Heat required = (2__0__lbs)(1.00__0__ BTU/lb Fº) (1.__0__ Fº)= 2__0__
BTU
(just *barely* 2 sig. figs.)

Also
2__0__ BTU = (20 BTU/1) (778ft lb/BTU) = 15,560 ft lb = 16,000 BTU
(keeping
2 sig. figs.)

6. P= 12__5__ W = 0.12__5__ KW
T= 8.__0__ H

W=
Pt = (0.12__5__
KW)(8.__0__ H) = 1.0__0__ KWH

Cost
= (1.0__0__ KWH)(9.__0__
cents/KWH) = 9.__0__ cents

7.
Doubling Time =70/P = 70/14 = 5yrs

1982
$1.__0__
1987 $2.__0__
1992 $4.__0__
1997 $8.__0__
2002 $16.__0__