22022(40.0 J)5.00 kg.(4.00 m/s)sKmv===Thus, the normal force is 2222()(5.0 kg) (9.8 m/s )(20 N)44.7 N45 N.yxFmgF=−=−=≈22. From the figure, one may write the kinetic energy (in units of J) as a function of xas 2040 20sKKxxSincexWKFx=Δ =⋅ΔG, the component of the force along the force along +xis /20N.xFdKdx−The normal force on the block is
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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.