KUEmvmgdiik=+ΔΔth122sincosθμθbgwherevi=14..m s Dividing by mass and rearranging, we obtain 20.13m.2 (sincos )ikvdg==+(b) Now that we know where on the incline it stops (d'= 0.13 + 0.55 = 0.68 m from the bottom), we can use Eq. 8-33 again (with W= 0 and now with Ki= 0) to describe the final kinetic energy (at the bottom): Emvmgdfk=−−=′−th122cosbgwhich — after dividing by the mass and rearranging — yields vgdk=′−=227cos..bgms (c) In part (a) it is clear that dincreases if μkdecreases — both mathematically (since it is a positive term in the denominator) and intuitively (less friction — less energy “lost”). In part (b), there are two terms in the expression for vwhich imply that it should increase if kwere smaller: the increased value of d'= d0+ dand that last factor sin
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