Question

A massless spring of spring constant k = 3717 N/m is connected to a mass m = 487 kg at rest on a horizontal, frictionless surface. A) The mass is displaced from equilibrium by A = 0.81 m along the spring's axis. How much potential energy, in joules, is stored in the spring as a result? B) When the mass is released from rest at the displacement A = 0.81 m, how much time, in seconds, is required for it to reach its maximum kinetic energy for the first time? 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 = 6.1178362 GJ. Calculate the number, N, of springs with spring constant k = 3717 N/m displaced to A = 0.81 m you would need to store 1 BOE of potential energy. D) Imagine that the N springs from part (c) are released from 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 N-spring system? E) Though not a practical system for energy storage, how many million buildings, B, each using 105 W, could the spring system temporarily power?

Answer 1

A massless spring of spring constant k = 3717 N/m is connected to a mass m = 487 kg at rest on a horizontal, frictionless surface. The potential energy stored in the spring when the mass is displaced by A = 0.81 m is 1214.692 Joules. The time required for the mass to reach its maximum kinetic energy is 0.362 seconds. You would need approximately 5.04 million springs to store 1 BOE of potential energy. The average power released by the N-spring system is approximately 3356.13 Watts. The spring system could temporarily power approximately 31.96 million buildings each using 105 W.

A) Potential Energy Stored in the Spring:

The potential energy stored in the spring is given by the formula:

PE = 1/2 k A^2

where k is the spring constant and A is the displacement.

PE = 1/2 x 3717 N/m x (0.81 m)^2

PE = 1/2 x 3717 x 0.6561

PE = 1214.692 Joules

So, the potential energy stored in the spring is 1214.692 Joules.

B) Time to Reach Maximum Kinetic Energy:

The time required for the mass to reach its maximum kinetic energy can be found using the formula:

T = sqrt(m/k)

T = sqrt(487 kg/3717 N/m)

T = sqrt(0.131 s^2)

T = 0.362 s

So, the time required for the mass to reach its maximum kinetic energy is 0.362 seconds.

C) Number of Springs for 1 BOE:

The number of springs needed to store 1 BOE of energy is calculated using the formula:

N = BOE energy / Energy per spring

N = 6.1178362 x 10^9 J / 1214.692 J/spring

N ≈ 5.04 x 10^6

So, you would need approximately 5.04 million springs to store 1 BOE of potential energy.

D) Average Power Released:

The average power released by the N-spring system can be calculated using the formula:

Power = Total Energy / Time

Since the potential energy is fully converted to kinetic energy, the total energy is the potential energy stored in the springs. Using the value found in part A (1214.692 Joules) and the time calculated in part B (0.362 seconds):

Power = 1214.692 J / 0.362 s

Power ≈ 3356.13 W

So, the average power released by the N-spring system is approximately 3356.13 Watts.

E) Number of Buildings Powered:

The number of buildings that can be powered by the spring system can be calculated by dividing the total power by the power used by each building:

Number of Buildings = Total Power / Power per Building

Given that each building uses 105 W and the total power is calculated in part D (3356.13 W):

Number of Buildings = 3356.13 W / 105 W/building

Number of Buildings ≈ 31.96

So, the spring system could temporarily power approximately 31.96 million buildings each using 105 W.

Answer 2

The potential energy stored in a displaced mass-spring system is 1219.98935 J. It takes approximately 1.13 seconds for the mass to reach its maximum kinetic energy. The number of springs required to store the energy equivalent of 1 barrel of crude oil is 5011991, and the spring system can temporarily power about 51547600 buildings using 105 W each.

Step-by-step explanation:

The student has asked several questions related to a mass-spring system on a frictionless surface.

A) Potential Energy Stored in the Spring

The potential energy (PE) stored in a spring when it is compressed or stretched by a distance A from its equilibrium position is given by the formula PE = 1/2 kA². Using the spring constant k = 3717 N/m and the displacement A = 0.81 m, the potential energy stored in the spring is:

PE = 1/2 × 3717 N/m × (0.81 m)² = 1/2 × 3717 × 0.6561 J = 1219.98935 J

B) Time to Maximum Kinetic Energy

Since no external forces are acting on the system, and it is on a frictionless surface, the maximum kinetic energy will occur at the equilibrium position, where all the potential energy has been converted to kinetic energy.

The oscillation period of a mass-spring system T is given by T = 2π√(m/k). Half of this period is the time required for the mass to reach maximum kinetic energy from the amplitude:

T/2 = π√(m/k) = π√(487 kg / 3717 N/m) = π√(0.1309611 s²) ≈ 1.13 s

C) Number of Springs to Store 1 BOE

1 BOE is equal to 6.1178362 GJ or 6.1178362 × 10⁹ J. The number of springs required to store this much energy N can be calculated by dividing the total energy by the energy stored in each spring:

N = (6.1178362 × 10⁹ J) / 1219.98935 J ≈ 5011991 springs

D) Average Power Released by N-Spring System

If we assume the energy is released in half the period of one oscillation (the time it takes for the mass to move from amplitude to equilibrium), the average power P in watts can be calculated by dividing the total energy by the time:

P = (6.1178362 × 10⁹ J) / (1.13 s) ≈ 5.4125076 × 10⁹ W

E) Number of Buildings Powered by the Spring System

If each building uses 105 W, the number of buildings B that can be powered temporarily by the spring system is:

B = (5.4125076 × 10⁹ W) / (105 W) ≈ 51547600 buildings