4x 105 j j the zero of potential energy is taken tthe

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Unformatted text preview: hat is her altitude? = -/2(210 J)/(1.4 kNfm) = 54.8 em. Problem J6tion .C@\measurethe altitude from sea level, U(y) #50m) = mg(y -1250 m) = -2.4x105 J, or 1'250 m - 2.4x105 Jj(GO kg}(9.8 O1/s2) = 842 m. ~./t',1/ 14. A more accurate expression for the force law of the rope in Example 8-3 is F = -kx + bx2 - cx3, where k and b have the values fven in Example 8-3, and c 3.1 N/m . Find the energy stored in stretching the rope 2.62 m. By what percentage does your result differ from that of Example 8-3? = 12S0m Solution J~. {l>lem >:l8cm? x Problem 10 Solution. The more accurate expression for force adds a term _ jl.62cm(_cx3)dx = = !(3.1 Njm3)x (2.62 01)4 = 36.5 J to the stored potential energy, which is about 4.9% of the 741 J of potential energy found in Example 8-3. Iicx41~.62m Problem 15. The force exerted by an unusual spring when it's compressed a distance x from equilibrium is given by F = -kx - cx3, where k = 220 N/m, and c = 3.6 kNjm3. Find the energy stored in this spring when it's been compressed 15 em. }...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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