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Chapter 4: Problem 1

\( \mathrm{A}\) mass of \(5 \mathrm{lbm}\) is released when it is just in contact with a spring of stiffness 25 lbf/ft that is attached to the ground. What is the maximum spring compression? Compare this to the deflection if the mass was just resting on the compressed spring. What would be the maximum spring compression if the mass was released from a distance of 5 ft above the top of the spring?

Short Answer

Expert verified
The spring compression is zero when the mass is just in contact with the spring. The spring compression is \(M \cdot g / k\) when the mass is just resting on the compressed spring. The maximum spring compression when the mass is released from a distance of 5 ft above the spring is \(x = \sqrt{(2 \cdot M \cdot g \cdot h) / k}\)

Step by step solution

01

Understanding the given quantities

The given quantities are the mass \(M = 5 \, \mathrm{lbm}\), the spring stiffness \(k = 25 \, \mathrm{lbf/ft}\), and the gravitational constant \(g = 32.2 \, \mathrm{ft/s^2}\). The weight of the object is \(W = M \cdot g\). The mass in slugs (the English unit of mass) is \(M \, \mathrm{(in \, slugs)} = \frac{M \, (in \, lbm)}{g}.\)
02

Calculate the maximum compression when the mass is just in contact with the spring

The energy balance equation can be set up as follows: the initial potential energy \(PE = W \cdot h\) equals the sum of the kinetic energy \(KE = \frac{1}{2} \cdot m \cdot v^2\) and the elastic potential energy \(EPE = \frac{1}{2} \cdot k \cdot x^2\). Since the object is in static equilibrium when it is just in contact with the spring, its initial potential energy and kinetic energy are zero, so the equation simplifies to \(M \cdot g \cdot 0 = 0 + \frac{1}{2}\cdot k \cdot x^2\). Solving this equation for \(x\) leads to \(x = 0 \, \mathrm{ft}\), i.e., the spring compression is zero when the mass is just in contact with the spring.
03

Calculate the spring compression if the mass was just resting on the compressed spring

For this case, the mass is in static equilibrium, so the spring force \(k \cdot x\) equals the weight of the object \(W = M \cdot g\). Solving this equation for \(x\) leads to \(x = \frac{M \cdot g}{k}\).
04

Maximum spring compression if the mass was released from a distance of 5 ft above the top of the spring

Now the initial potential energy \(PE = M \cdot g \cdot h\) equals the sum of the kinetic energy \(KE = \frac{1}{2} \cdot m \cdot v^2 = 0\) and the elastic potential energy \(EPE = \frac{1}{2} k \cdot x^2\). The initial potential energy is now nonzero because the object is released from a height \(h = 5 \, \mathrm{ft}\) above the spring. The kinetic energy is zero because the object is released from rest. Setting up the equation gives \(M \cdot g \cdot h = 0 + \frac{1}{2}\cdot k \cdot x^2\). Solve this equation for \(x\) to find the maximum spring compression.

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