Unformatted text preview: sses), y2 = For + mgy. (This is Equation 8-7 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 loop-the-loop " track shown...
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