ch09-p086

# Ch09-p086 - 86 From mechanical energy conservation(or simply using Eq 2-16 with a = g downward we obtain v = 2 gh = 2(9.8 m/s 2(1.5 m = 5.4 m/s for

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86. From mechanical energy conservation (or simply using Eq. 2-16 with G ag = downward) we obtain 2 2 2(9.8 m/s )(1.5 m) 5.4 m/s vg h == = 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 distance d = 0.30 m. Choosing + y downward, the deceleration a is found using Eq. 2-16. 02 2 54 2030 2 22 =+ ¡ =− va d a v d . (. ) which yields a 49 m s 2 . Thus, the magnitude of the net (vertical) force is m|a| = 49 m in SI units, which (since 49 m/s 2 = 5(9.8 m/s 2 ) = 5 g ) can be expressed as 5 mg . (b) During the deceleration process, the forces on the dinosaur are (in the vertical direction) N F G and mg G . If we choose + y upward, 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.

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