This preview shows page 1. Sign up to view the full content.
Unformatted text preview: sses), y2 = For + mgy. (This is Equation 87 for points A and B in the .... sketch.) Since the gravitational potential energy . hange is negligible compared to the other terms, k ':::! c .. m(vB/y)2 = 103 kg(2.4 km/s=15 m)2 ~ 25.6 MN/~. (Note: The surface gravity on the moon 18 1.62 m/s , o the maximum change in potential energy of a s packet is only mgy = 24.3kJ, while its kinetic energy, lmv~ = 2.88 OJ, is more than 106 times larger. likewise, the gravitational potential energy of the spring is negligible.) !k !mv1 Solution
The maximum energy stored by thesprings,~kx2, ;; without incurring damage, is equal to the maximum kinetic energy,!mv2, of the car before a collision. (We assume a level road and horizontal collision so that there is no change in the car's gravitational potential energy.) Thus, ~kX2 == ~mv2, or v = Jk/m x = " J(7.0xl05 N/m)/(1400 kg)(O.16 m) = 3.58 m/s == i" 12.9 km/h.
i: Problem , 32. A block slides on the frictionless looptheloop " track shown...
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