5.5. Applications from Physics, Engineering, and Statistics http://www.ck12.org
The graph ofF(x) = 80 − 2. 5 xrepresents the variation of the force with heightx(Figure 23). The work done
corresponds to computing the area under the force graph.
Figure 23
Thus the work done is
W=
∫b
a F(x)dx
=∫ 32
0 [^80 −^2.^5 x]dx
=[
80 x−^22.^5 x^2] 32
0
=1280 J.3.The rope alone.Since the weight of the rope is 5.1 N/m and the height is 32 meters , the total weight of the rope
from the floor to a height of 32 meters is
( 5. 1 )( 32 ) = 163 .2 N.But since the worker is constantly pulling the rope, the rope’s length is decreasing at a constant rate and thus its
weight is also decreasing as the bucket being lifted. So atxmeters, the( 32 −x)meters there remain to be lifted of
weightF(x) = ( 5. 1 )( 32 −x)N. Thus the work done to lift the weight of the rope is
W=
∫ 32
0 F(x)dx=∫ 35
0 (^5.^1 )(^32 −x)dx
W= ( 5. 1 )[
32 x−x2
2] 32
0
= 2611 .2 J.