86. From mechanical energy conservation (or simply using Eq. 2-16 with Gag=downward) we obtain222(9.8 m/s )(1.5 m)5.4 m/svgh===for the speed just as the body makes contact with the ground. (a) During the compression of the body, the center of mass must decelerate over a distanced= 0.30 m. Choosing +ydownward, the deceleration ais found using Eq. 2-16.022542030222=+¡=−vadavd.(. )which yields a49 m s2. Thus, the magnitude of the net (vertical) force is m|a|= 49min SI units, which (since 49 m/s2= 5(9.8 m/s2) = 5g) can be expressed as 5mg.(b) During the deceleration process, the forces on the dinosaur are (in the vertical direction)NFGand mgG. If we choose +yupward, and use the final result from part (a), we
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.