A massless spring of spring constant k = 4170 N/m is connected to a mass m = 13

Question

A massless spring of spring constant k = 4170 N/m is connected to a mass m = 13 kg at rest on a horizontal, frictionless surface.

20% Part (a) The mass is displaced from equilibrium by A = 0.69 m along the springs axis. How much potential energy in Joules, is stored in the spring as a result? ? 20% Part (b) when the mass is released from rest at the displacement.4= 0.69 m,how much time, in seconds, is required for it to reach its maximum kinetic energy for the first time? 20% Part (c) The typical amount of energy released when burning one barrel of crude oil is called the barrel of oil equivalent (BOE) and is equal to 1 BOE = 6.11 78362 GJ. Calculate the number, N of springs with spring constant k = 4170 Nm displaced to.?0 69 m you would need to store 1 BOE of potential energy. ? 20% Part (d) Imagine that the springs from part (c) are released trom rest simultaneously. If the potential energy 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 energy is released? That is, what would be the average power, in watts, released by the Nspring system? -.. 20% Part (e) Though not a practical system for energy storage, how many million buildings, B each using 105 W could the spring system temporarily power?

Explanation / Answer

a) potential energy = 0.5kx^2

= 0.5*4170*0.69^2

= 992.66 J

b) time period = 2pi*sqrt(m/K)

T = 2pi*sqrt(13 / 4170)

= 0.35 s

so time taken to reach maximum KE = T/2 = 0.35 / 2 = 0.175 s

c) number of springs required = 6.1178362*10^9 / 992.66 = 6*10^6

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