Question

Hooke's Law In Exercises 3-10, use Hooke's Law to determine the variable force in the spring problem. A force of 5 pounds compresses a 15 -inch spring a total of 4 inches. How much work is done in compressing the spring 7 inches?

Short answer

The work done in compressing the spring 7 inches is 20.375 pounds*inches.

Step-by-step solution

Find the Spring Constant 'k'

We are given that the force exerted on the spring is 5 pounds and this compresses the spring by 4 inches. Using Hooke's Law \(F=-kx\), replacing the force F with 5 pounds and the displacement x with 4 inches, we have \(5=-k*4\). By isolating 'k', we find that the spring constant \(k=-5/4 pounds/inche or k =-1.25 pounds/inch\). The spring constant is negative because it's a restoring force, meaning it acts in the opposite direction to the displacement.

Calculate the Work Done

Work done on a spring is given by the integral of the force from the initial to the final displacement. The initial compression was 4 inches, and we need to find the work done to compress it a total of 7 inches. So, we set the integral bounds from 4 to 7. The force is given by Hooke's Law \(F=-kx\). Substituting the value we found for 'k' into the force equation, we get \(F=-(-1.25x) = 1.25x\). Then, work \(W = \int _{4}^{7}(1.25x)dx = [0.625x^{2}]_{4}^{7}=(0.625*49)-(0.625*16)=20.375 pounds*inches\). The work is positive because it's energy spent in further compressing the spring.