**Unformatted text preview: **less), then the total mechanical energy of the masses is conserved. Each mass has gravitational potential energy mgy, measured from zero on the floor, and kinetic energy!mv2 Since the masses are connected (by a string of presumably fixed length), their velocities have equal magnitudes when both masses are off the floor. (a) If we take position A when the 4-kg mass is on the floor at rest, and position Bwhen the 7-kgmass is about tostrike the floor, then the conservation of energy requires UA + KA = UB + KB, or (7x9.8 N)(5 m) = !(7 kg)vb + !(4 kg)vb + (4x9.8 N)(5 m). Thus, vs= J2(3x9.8 N)(5 m)/(l1 kg) = 5.17 mls. (b) The 4-kg mass, with an upward velocity of 2kg , , FIGURE 8-31 Problem 35. '; Solution
Since friction is absent by assumption, mehhanical energy is conserved, and Uo + Ko = U1 + 1<1, where 1; tL"
- CHAPTER 8
will rise (as a projectile) an additional 131 =(5,17 m/s)2/(2x9.8 . ilium height of 5 m + 1.36 m m/s ) = 1.36 m to a 2 = 6.36 m off the
i; /(Note: the mechanical energy of both masses is
gerconserved after the 7-kg mass hits t...

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