Question

(10%) Problem 7: A massless spring of spring constant k = 5701 Nm is connected to mass m = 123 kg at rest 0n horizontal frictionless surface 20% Part (a) The mass is displaced from equilibrium by A = 0.59 m along the spring axis. How much potential energy: in joules. is stored in the spring as result? 20% Part (b) When the mass is released from rest at the displacement 4= 0.59 m how much time. in seconds_ is required for it to reach its maximm 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 BOE 6.1178362 GJ. Calculate the number; N; of springs with spring constant k = 5701 Nim displaced to 4 0.59 m YOu would need to store BOE of potential energy: 20% Part (d) Imagine that the Vsprings 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 what would be the average power; in watts. released by the Nspring system? 20% Part (e) Though not practical system for energy storage: how many million buildings_ each using 105 W. could the spring system temporarily power?

Answer 1

Final Answers:

(a) Potential energy stored in the spring is 508.144 J.

(b) Time required to reach maximum kinetic energy is 1.2 s.

(c) Number of springs needed to store 6.1178362 GJ is approximately 12,025.

(d) Average power released by the spring system is 4.01 kW.

(e) The spring system could temporarily power around 38,095,238 buildings.

Explanations:

(a) The potential energy stored in a spring is given by the formula
`\( U = (1)/(2)kA^2 \)` where
`\( k \)` is the spring constant and
`\( A \)` is the displacement. Substituting the values, we get
`\( U = (1)/(2) * 5701 * (0.59)^2 \)` resulting in
`\( U = 508.144 \)` Joules.

(b) The time required to reach maximum kinetic energy for a mass-spring system released from rest is given by the formula
`\( T = \sqrt{(m)/(k)} \)` where
`\( m \)` is the mass and
`\( k \)` is the spring constant. Substituting the values
`\( T = \sqrt{(123)/(5701)} \)` yielding
`\( T = 1.2 \)` seconds.

(c) The number of springs needed to store a given amount of potential energy is determined by dividing the total energy by the energy stored in one spring. Thus
`\( N = (6.1178362 * 10^9)/(508.144) \approx 12,025 \)`.

(d) The average power released by the spring system, assuming full conversion from potential to kinetic energy is equal to the potential energy divided by the time taken. Using
`\( P = (U)/(T) \)` where
`\( U \)` is potential energy and
`\( T \) is time, we find \( P = (508.144)/(1.2) \approx 4.01 \) kilowatts.`

(e) To determine the number of buildings the spring system can power, we divide the total power by the power used by each building:
`\( (4.01 * 10^3)/(10^5) \)` yielding approximately 38,095,238 buildings.

Answer 2

The potential energy stored in the spring is 988.2879 joules. The time required to reach maximum kinetic energy is 0.3305 seconds. The spring system can theoretically power approximately 113 million buildings each using 105 W.

Step-by-step explanation:

To solve the problem, we will use principles of energy conservation and simple harmonic motion from physics.

Part (a)

The potential energy (PE) stored in a spring is given by the formula PE = 0.5× k ×A2, where k is the spring constant and A is the displacement. Inserting the given values, PE = 0.5 × 5701 N/m × (0.59 m)2 = 988.2879 J.

Part (b)

The time to reach maximum kinetic energy for the first time is a quarter of the period (T) of the oscillation. The period is given by T = 2 × π × √(m/k), so time = T/4. Therefore, time = (π × √(123 kg / 5701 N/m))/2 = 0.3305 seconds.

Part (c)

To find the number N of springs needed to store BOE of potential energy, N = BOE/PE of one spring. With BOE = 6.1178362× 109 J and PE of one spring = 988.2879 J, N = 6.19 × 106.

Part (d)

The average power is the energy divided by the time taken to release it, which is a quarter of the period. Thus, power = N ×PE / (T/4), which gives us a power of 1.19× 1010 watts.

Part (e)

Considering that each building uses 105 W of power, the number of buildings powered = Power available from springs / Power per building = (1.19 × 1010 W) / (105 W) = 1.13 ×108 buildings or 113 million buildings approximately.