(7%) Problem 10: A massless spring of spring constant-8501 Nm is connected to a
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(7%) Problem 10: A massless spring of spring constant-8501 Nm is connected to a mass m-377 kg at rest on a horizontal. 2000 Part (a) The mass is displaced from equiliorium by d lre he spring as a resul? 0.97 m along the spring's axis. Ilow much potential energy n joules, is Grade Summary Potcntial Submissions PE = 00% % pet allermui detailed view atan acota sinh cosh tanotano END Degrees Radians ULCLEAR Submit Ilint I zive up! llints: 0% deduction per hint 11-....remaining: 1 Teedback: 0% deduction per feedback. 2010 Part (b) when the mass is released from rest at the displacement. 4-0.97 m. how much time, in seconds, is required for it to reach its maximum kinctic cncrgy for the first time? 2096 Part (c) The typical amount of crorgy released when burning one barel of crude oil is callcc tho barrel of oil cquivalent (BOE) and is equal to 1 BOE 6.1173362 GJ. Calculate the number, N, of springs with spring constant k 8501 N/m displaced toA -7myou would nced to store BOE of potential cncrgy 2096 Part (d) Imaginc that the N springs from part (c) are released from rest simu-tancous-v If thc potential cnergy stored in the springs is fully converted to kinetic energy and thereby "released" when the attached masses pass through equilibrium, what would be the average rate at which the cncrgy is relcased? That is, what would be the average power in wats, relcascd by the Nspring system? 20% Part (e) 1 hough not a practical system for energy storage, how many million buildings, s, each using 105 W could the spring system temporarily power?Explanation / Answer
a)
Es = spring potential energy = (0.5) k A2 = (0.5) (8501) (0.97)2 = 4000 J
b)
T = time period = 2pi sqrt(m/k) = (2 x 3.14) sqrt(377/8501) = 1.32 sec
t = time required = T/4 = 1.32/4 = 0.33 sec
c)
N = 6.1178362 x 109 /((0.5) k A2) = 6.1178362 x 109 /(4000) = 1.53 x 106
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