This preview shows page 1. Sign up to view the full content.
Unformatted text preview: The gravitational force on an object is constant, F 9 -mgj, while the paths are (a) dr=jdy for x = 0 and w= fF'dr=
+ laF(Y=O).idx+ l
0 aF (x=a)'jdY = 1 0 F(y = a).idx+ 1 F(x = 0) .jdy 124 CHAPTER 8 (Note that a path parallel to the x-axis ftom right to left is represented by dr= ida; with x goihgfrom a to in the limits of integration, etc.) (a) Fdi: F = Fo,j, the expression for the work becomes II Solution
To avoid confusion, let x-y refer to the Earth's surface, and x' ~y' refer to the incline. The gravitational potential energy (g assumed constant, zero at y 0) is U = mgy. For a point on the incline, y x' sinO, so U mgx'sinO. o W = O+Fo ~ r dy+O+ Fo 1
a dy dy = Fda- Foa = O. I
I! = = = (b) For F= Fo(xja)j,
W the work around tije square is = 0 + Fa (~) l a ~I
~ +0 +0 \
\ Section 8-2: Problem Potential Energy .
:! 4. Rework Example 8-1, now taking potential energy at street level. of Problem 6 Solution. Solution .; Problem (a) The office in question is 32 stories ab~~e the street level (the first floor) where Ul 0, so thel!difference in gravitational potential en...
View Full Document