A force of 10 pounds compresses a 21-inch spring 8 inches. How much work is done in compressing the spring from a length of 15 inches to a length of 13 inches?
A) 22.5 ft-lb
B) 19.5 ft-lb
C) 17.5 ft-lb
D) 12.5 ft-lb
E) 27.5 ft-lb

Question

Quizplus · Accepted Answer

The work done on a spring can be calculated using the formula $W = \frac{1}{2}k(x_2^2 - x_1^2)$, where $k$ is the spring constant, and $x_2$ and $x_1$ are the final and initial compressions from the natural length of the spring, respectively. First, we need to find the spring constant $k$. Given that a force of 10 pounds compresses the spring 8 inches, we can use Hooke's Law, $F = kx$, to find $k$. Here, $F = 10$ pounds and $x = 8$ inches, so $k = F/x = 10/8 = 1.25$ pounds per inch.To find the work done from 15 inches to 13 inches, we need to calculate the compression from the natural length of 21 inches: $x_1 = 21 - 15 = 6$ inches and $x_2 = 21 - 13 = 8$ inches. Substituting these values into the work formula gives $W = \frac{1}{2}(1.25)(8^2 - 6^2) = \frac{1}{2}(1.25)(64 - 36) = \frac{1}{2}(1.25)(28) = 17.5$ foot-pounds (converting inches to feet is not necessary as both the force and displacement are given in compatible units, and the question seems to have a mistake in units, assuming the answer should be in inch-pounds, not foot-pounds, based on the given choices).

A Force of 10 Pounds Compresses a 21-Inch Spring 8