Physics 1110                Problem Set 1

 

Note:    Include units in all problems

 

A. Write the following numbers in ordinary notation:

1)      105 cm                                     3) 2.56 x 108 ft3

2)      10-5 sec                                                4) 73.89 x 109 BTU’s

 

B. Express 55mi/H in km/H

 

C. Express 720 in2 in ft2

 

D. Express 6048 in3 in ft3

 

E. At a cost of $5.50 per square yard of material, how much will a person have to pay for a strip of 

     material 4 feet wide by 20 feet long?

 

F. A water faucet has a cross sectional area of 1.5 in2. Suppose water flows from the faucet at a speed of

    6 in/sec. What volume of water flowing from the faucet during every second?

 

G. A horizontal force of 10 lbs. is exerted on a block in sliding it along a horizontal surface a distance of

    7.2 ft. How much work has been done by the force?

 

H. A vertical force of 4.5 Newtons is applied to a crate in order to lift it to a table 1.2 m high. How much

     work has been done by the force?

 

 

Physics 1110                                        Solution to Problem Set 1

(Note: Please do not look at this until you have attempted all of the previous problems in Problem Set 1.)

 

A.        1) 100,000cm              3) 256,000,000 ft3         

            2) 0.00001 sec             4) 73,890,000,000 BTU’s

 

B.                 1.6 km = 1mi, hence (55 mi/H)(1.6 km/1 mi) =88 km/H

 

C.                 144 in2 = 1 ft2, hence 720 in2 (1 ft2/ 144 in2) = 5 ft2

 

D.                 1728 in3 = 1 ft3, hence 6048 in3 (1 ft3/1728 in3) = 3.5 ft3

 

E.         The area of the material is 4ft x 20 ft =80ft2

            However, 3 ft = 1yd or 9ft2 = 1yd2, hence 80 ft2 (1yd2/9ft2)($5.50/yd2) = $48.89

 

F.                  This particular problem we have not determined how to do in class. However, one can deduce how to work the problem from what is asked for in the problem—i.e. what volume (in3) is flowing per second(s)? Indeed the volume rate of flow can be determined by multiplying the speed (or velocity) times the area. Hence (6 in/sec)(1.5 in2) = 9in3/sec

 

G.                 W =Force x distance = (10 lbs)(7.2 ft) = 72 ft. lbs.

 

H.        W= Force x distance = (4.5 N)(1.2m) = 5.4 joules