ch08-p060

# ch08-p060 - 60. This can be worked entirely by the methods...

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KU E mv mgd i ik =+ ΔΔ th 1 2 2 sin cos θμ θ bg where v i = 14 .. m s Dividing by mass and rearranging, we obtain 2 0.13m. 2 (sin cos ) i k v d g == + (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 K i = 0) to describe the final kinetic energy (at the bottom): E mv mgd f k =− = th 1 2 2 cos b g which — after dividing by the mass and rearranging — yields vg d k = −= 22 7 cos . . b g ms (c) In part (a) it is clear that d increases if μ k decreases — 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 v which imply that it should increase if k were smaller: the increased value of d' = d 0 + d and that last factor sin
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