This preview shows page 1. Sign up to view the full content.
Unformatted text preview: e equivalent to a maSs loss of dm/dt = d(E/Cl)/dt = P/c2 = (3.85x1026 J/s) + (3xl08 m/s)2 = 4.28x109 kg/so (This is only about 7xlO14 solar masses per year.) Paired Problems
'I FIGURE 843 Problem 66 Solution. Problem
65. A block slides down a frictionless incline that terminates in a ramp pointing up at a 450 angle, as shown in Fig. 842. Find an expression for the horizontal range x shown in the figure, as a function ofthe heights hI and h2 shown." the end of the ramp. Now there is pote:ntialenergy of the spring as well as gravity. If point 1 is taken to be the block's position at maximum compression of the spring; then !)'K = !mv~ 0 = !).U = !kd2  mgh. Thus x = (kd2/mg)  2h. Problem
67. A ball of mass m is being whirled around on a string of length R in a vertical circle; the string does no work on the balL (a) Show from force considerations that the speed at the top of the circle must be at least v'R9 if the string is to remain taut. (b) Show that, as long as the string remains taut, the speed at the bottom of the circle can be no more than v'5 times the speed at the top. Solution
After l...
View
Full
Document
This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.
 Spring '08
 WORMER
 Physics, Conservation Of Energy, Energy, Work

Click to edit the document details