ch08-p066 - 66(a Since the speed of the crate of mass m...

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22 11 (300kg)(120m/s) 216J. Km v == = (b) The magnitude of the kinetic frictional force is 23 (0.400)(300kg)(9.8m/s ) 1.18 10 N. N fF m g μμ = = × (c) Let the distance the crate moved relative to the conveyor belt before it stops slipping be d , then from Eq. 2-16 ( v 2 = 2 ad = 2( f / m ) d ) we find Δ Ef dm v K th = 1 2 2 . Thus, the total energy that must be supplied by the motor is th 2 (2)(216J) J. WK E K = + Δ = = = 432 (d) The energy supplied by the motor is the work W it 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.

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