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Chapter 13: Problem 31

A 5 -kg collar \(A\) is at rest on top of, but not attached to, a spring with stiffiness \(k_{1}=400 \mathrm{N} / \mathrm{m}\) when a constant \(150-\mathrm{N}\) force is applied to the cable. Knowing \(A\) has a speed of \(1 \mathrm{m} / \mathrm{s}\) when the upper spring is compressed \(75 \mathrm{mm}\), determine the spring stiffness \(k_{2} .\) Ignore friction and the mass of the pulley.

Short Answer

Expert verified
The spring stiffness \( k_2 \) is approximately 2711.11 N/m.

Step by step solution

01

Identify Known Quantities

We have a collar of mass, \( m = 5 \text{ kg} \), with an initial velocity of \( 0 \text{ m/s} \) at rest. The spring stiffness \( k_1 = 400 \text{ N/m} \), a constant force \( F = 150 \text{ N} \) is applied, and at a speed of \( 1 \text{ m/s} \), spring \( k_1 \) is compressed by \( x_1 = 75 \text{ mm} = 0.075 \text{ m} \). We need to find the spring stiffness \( k_2 \).
02

Set Up the Work-Energy Principle

We use the work-energy principle, which states that the work done by external forces and potential energy changes (spring compression) equal the change in kinetic energy:\[ W - \Delta U = \Delta KE \]Here, \( W \) is the work done by the applied force, \( \Delta U \) is the change in potential energy due to spring compression, and \( \Delta KE \) is the change in kinetic energy.
03

Compute the Work Done by the Force

The work done \( W \) by the constant force \( F \) is \( W = F \cdot x \), where \( x \) is the displacement of the point where force is applied. Since the speed is given when spring \( k_1 \) is compressed by \( 0.075 \text{ m} \), work done is:\[ W = 150 \text{ N} \times 0.075 \text{ m} = 11.25 \text{ J} \]
04

Calculate Change in Potential Energy of Springs

The change in potential energy due to the first spring \( k_1 \) is:\[ \Delta U_{k_1} = \frac{1}{2}k_1x_1^2 = \frac{1}{2} \times 400 \text{ N/m} \times (0.075 \text{ m})^2 = 1.125 \text{ J} \]Let the change in potential energy of spring \( k_2 \) be \( \Delta U_{k_2} = \frac{1}{2}k_2x_2^2 \). The displacement for this spring is also \( 0.075 \text{ m} \), so we have:\[ \Delta U_{k_2} = \frac{1}{2}k_2(0.075)^2 \]
05

Calculate Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) when speed is \( 1 \text{ m/s} \) is:\[ \Delta KE = \frac{1}{2}m(v^2 - u^2) = \frac{1}{2} \times 5 \times ((1)^2 - (0)^2) = 2.5 \text{ J} \]
06

Formulate the Equation

Using the work-energy principle equation:\[ W - (\Delta U_{k_1} + \Delta U_{k_2}) = \Delta KE \]Substitute the values:\[ 11.25 - (1.125 + \frac{1}{2}k_2(0.075)^2) = 2.5 \]
07

Solve for Spring Stiffness \( k_2 \)

Simplify the equation:\[ 11.25 - 1.125 - \frac{1}{2}k_2(0.075)^2 = 2.5 \]\[ \frac{1}{2}k_2(0.075)^2 = 11.25 - 1.125 - 2.5 \]\[ \frac{1}{2}k_2(0.075)^2 = 7.625 \]Multiply through by 2 and solve for \( k_2 \):\[ k_2(0.075)^2 = 15.25 \]\[ k_2 = \frac{15.25}{(0.075)^2} \approx 2711.11 \text{ N/m} \]
08

Final Result

Thus, the stiffness \( k_2 \) of the second spring is approximately \( 2711.11 \text{ N/m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Stiffness Calculation
The spring stiffness, often denoted as "k," is an essential parameter that measures how resistant a spring is to being compressed or stretched. Knowing how to calculate spring stiffness is crucial in solving problems involving springs.

In our exercise, we deal with two springs with the first having a known stiffness (\(k_1 = 400\, \mathrm{N/m}\)) and the other unknown spring stiffness (\(k_2\)). The force applied to compress the springs and the displacement caused by this force helps us determine the unknown stiffness.To calculate spring stiffness:
  • Determine the change in potential energy due to spring compression.
  • Use the work-energy principle which relates the work done by external forces to these changes in energy.
  • Utilize the equation: \[\Delta U_{\text{spring}} = \frac{1}{2} k x^2\]where \(x\) is the displacement of the spring.
By rearranging terms in the work-energy equation, you can isolate \(k_2\) to solve for the spring stiffness of the second spring.
Potential Energy Change
Potential energy in the context of springs is related to the position of an object under the effect of elastic forces. When a spring is either compressed or stretched, it stores energy. This stored energy is called elastic potential energy.

In the exercise, the potential energy change occurs when the spring is compressed by a certain distance. The change in potential energy (\(\Delta U\)) can be calculated using:\[\Delta U = \frac{1}{2} k x^2\]
  • \(k\) is the spring stiffness (either \(k_1\) or \(k_2\) depending on the spring in use).
  • \(x\) is the compression or extension distance.
This formula allows us to quantify how much energy is either stored in or released from a spring when it's moved from its natural length. It is a critical component of the work-energy principle, helping us understand the interplay between forces and energy.
Kinetic Energy Change
Kinetic energy is the energy an object possesses due to its motion. When examining movement in systems like the one in the exercise, calculating the change in kinetic energy gives insight into how motion influences energy changes in a system.

Kinetic energy (\(KE\)) is calculated using:\[KE = \frac{1}{2} m v^2\]
  • \(m\) is the mass of the object.
  • \(v\) is the velocity.
In our scenario, the collar starts at rest and gains speed due to an external force, altering its kinetic energy. The change in kinetic energy (\(\Delta KE\)) is important because it links directly to the work done by the external force and the potential energy exchanged as the spring compresses:\[\Delta KE = \frac{1}{2} m (v^2 - u^2)\]Here \(u\) is the initial velocity (0 for a resting object), and \(v\) is the final velocity. By evaluating the change in kinetic energy, we learn how effectively the system's energy has been converted into motion.

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Most popular questions from this chapter

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