KUEmgdfk=−−=−''th(sincos )θμwhich leads to the speed at the bottom of the ramp vKmgdfk==−=2255ms.sincos.θμbg(b) This speed begins its horizontal motion, where fk= kmgand ΔU= 0. It slides a distanced'before it stops. According to Eq. 8-31 (with W= 0), 001201222=++++′−+′ΔΔKUEmvmgdgdgdkkkΔthθ μsincosbgchwhere we have divided by mass and substituted from part (a) in the last step. Therefore, ′ =−=ddkkcos..bg54m (c) We see from the algebraic form of the results, above, that the answers do not depend on mass. A 90 kg crate should have the same speed at the bottom and sliding distance across the floor, to the extent that the friction relations in Ch. 6 are accurate. Interestingly, sincegdoes not appear in the relation for d', the sliding distance would seem to be the
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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.