2211(300kg)(120m/s)216J.Kmv===(b) The magnitude of the kinetic frictional force is 23(0.400)(300kg)(9.8m/s ) 1.18 10 N.NfFmgμμ==×(c) Let the distance the crate moved relative to the conveyor belt before it stops slipping bed, then from Eq. 2-16 (v2= 2ad= 2(f/m)d) we find ΔEfdmvKth=122.Thus, the total energy that must be supplied by the motor is th2(2)(216J)J.WK EK=+ Δ=== 432(d) The energy supplied by the motor is the work Wit does on the system, and must be
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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.