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# Force with potential energy u mgy above a reference

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Unformatted text preview: ndicular to the displacement along the track (by definition); hence it does no work on the particle, and the mechanical energy is conserved: ~mv2 + mgy = constant. Since the height y = ax2 is given in terms of the horizontal displacement, we may write !mv2 + mgax2 = constant, or v2 + 2gax2 = (a different) constant. One can see that the maximum speed occurs. when the displacement is a minimum (zero, in fact) and vice-versa. Thus, One can determine the constant two ways: v2 + 2gax2 = v~ax = 2gax~ax' The turning points are the places where the velocity is instantaneously zero (a minimum) and hence are given by x~ax = :t.vv'fnax/2ga = Potential Energy Curves iparticle slides along the frictionless track shown ~ig. 8-34, starting at rest from point A. Find ) its speed at B, (b) its speed at Cl and (c) the proximate location of its right-hand turning 6fut~ frbn 2 and icle has kinetic energy ll.t1(>nalpotential energy mgy (measured .above !mv II Jl 132 CHAPTER 8 -:t:.J(8.5 m/s)2/2(9:8 mjs2)(O.92 m-I) =:1:2.00 m. (In this case; the particle's motion is oscillatory, b...
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