Problem 49 - Calculus Early Transcendentals

Question

Hooke's law is applicable to idealized (linear) springs that are not stretched or compressed too far. Consider a nonlinear spring whose restoring force is given by \(F(x)=16 x-0.1 x^{3},\) for \(|x| \leq 7\) a. Graph the restoring force and interpret it. b. How much work is done in stretching the spring from its equilibrium position \((x=0)\) to \(x=1.5 ?\) c. How much work is done in compressing the spring from its equilibrium position \((x=0)\) to \(x=-2 ?\)

Step-by-Step Solution

Answer

In summary, we analyzed a nonlinear spring system with a given restoring force function \(F(x) = 16x - 0.1x^3\). After graphing the function, we found that the force always tries to bring the mass back to its equilibrium position. Then, we calculated the work done to stretch the spring from its equilibrium position to \(x = 1.5\) and found it to be 18 Joules. Finally, we calculated the work done to compress the spring from its equilibrium position to \(x = -2\) and found it to be 42.4 Joules. The graph and calculations demonstrate how the restoring force varies with the spring's displacement from its equilibrium position and helps us to quantify the work required for stretching or compressing the spring.

Step 1: Part a - Graphing the Restoring Force

To graph the function \(F(x) = 16x - 0.1x^3\), we can plot it on a graphing tool (like Desmos or a graphing calculator). The graph will show the restoring force of the spring acting in both positive and negative directions of motion. Since we are dealing with a nonlinear spring, the graph will not be a straight line, but a curve. Please plot the function \(F(x) = 16x - 0.1x^3\) and observe its behavior. It's important to notice that, for \(x>0\), the force is positive, and for \(x<0\), the force is negative. It means that the force always tries to bring the mass (attached to the spring) back to its equilibrium position (when \(x=0\), \(F_x=0\)).

Step 2: Part b - Work Done in Stretching the Spring

To find the work done in stretching the spring from \(x = 0\) to \(x = 1.5\), we'll calculate the integral of the restoring force function \(F(x)\) with respect to \(x\) over the given interval. Using the work formula: \(W = \int_a^b F(x) dx\) In this case, \(a=0\) and \(b=1.5\). Therefore, \(W = \int_0^{1.5} (16x - 0.1x^3) dx\) Now, calculate the integral and find the work done: \(W = \left[8x^2 - 0.025x^4\right]_0^{1.5}\) \(W = (8(1.5)^2 - 0.025(1.5)^4) - (8(0)^2 - 0.025(0)^4)\) \(W = 18 J\) So, the work done in stretching the spring from its equilibrium position to \(x = 1.5\) is 18 Joules.

Step 3: Part c - Work Done in Compressing the Spring

To find the work done in compressing the spring from \(x = 0\) to \(x = -2\), we'll calculate the integral of the restoring force function \(F(x)\) with respect to \(x\) over the new interval. Using the work formula: \(W = \int_a^b F(x) dx\) In this case, \(a=0\) and \(b=-2\). Therefore, \(W = \int_0^{-2} (16x - 0.1x^3) dx\) Now, calculate the integral and find the work done: \(W = \left[8x^2 - 0.025x^4\right]_0^{-2}\) \(W = (8(-2)^2 - 0.025(-2)^4) - (8(0)^2 - 0.025(0)^4)\) \(W = 42.4 J\) So, the work done in compressing the spring from its equilibrium position to \(x = -2\) is 42.4 Joules.