Solutions for Problem Set 7

Last revised 1999/10/18
HRW pp. 149-154 # 24, 27, 29, 31, 40, 50, sheet # XII.

Further examples: HRW pp. 149-154 # 36, 53.

24.


27. Count the squares between the plot of the function and the force=0 line in the graph in HRW. There are 2 + 1 + 0 - 1/2 = 2.5 blocks; the minus sign in the last number is because the function becomes negative between 6 and 8 m. Since each block is 5 Newtons high and 2 meters wide,

W = sum( F d ) = 2.5 * 5 N * 2 m
W = 25 Joules

29.
(a) Obviously, there should be 12 blocks under the curve, each 1/2 N m = 1/2 J in size. The closest I could get was 11 3/4.

(b) Since the angle between the force and the displacement is 0 degrees, the work done is given by

W = integral ( F dx )

W = integral ( a / x2 dx)

W = ( - 1) a / x

where the integral runs from 1 to 3 and a = 9 N m2 . Hence

W = (-1) (9 N m2) [ (1 / 3 ) - (1 / 1) ]

W = 6 Joules.

31.

36.

40. Use HRW figure 7-41 for the diagram.
(From MATHCAD file)






50.

Given:

P1, the power required at v1, = 10 hp
v1 = 2.5 mi/hr
v2 = 7.5 mi/hr

Find:

P2, the power required at v2

It is also stated that the force is proportional to velocity, so that for some constant a, F = a v

Relations:
Definitions of power and work:
P = d W / dt
W = F x

Combining these two

P = d (F x) / dt = F v

Hence using the expression for F

P = ( a v ) v = a v2

Taking ratios

P1 / P2 = ( v1 / v2 )2

P2 = P1 ( v2 / v1 )2

P2 = 9 P 1 = 90 hp

regardless of the size of a horsepower or the value of the constant a.

53. (From MATHCAD file)



XII. (From MATHCAD file)


The area under the curve is about 15.5 squares x 0.8 j/square = 12.4 j

Links to image files used in these solutions:

24, 24a, 24b, 24c, 24d.
31a, 31b.
36a, 36b.
40a, 40b, 40c, 40d, 40e.
53a, 53b, 53c.
mvb12
mvb12a