Solve the problem.
-The spring of a spring balance is 5.0 in. long when there is no weight on the balance, and it is 7.8 in. long with 4.0 lb hung from the balance. How much work is done in stretching it from 5.0 in. to a length of 13.3 in.?
A) 49 in.-lb
B) 24 in.-lb
C) 110 in.-lb
D) 5.9 in.-lb

Question

Quizplus · Accepted Answer

The work done in stretching 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, $x_2$ is the final length, and $x_1$ is the initial length. First, we need to find the spring constant $k$. The extension from 5.0 in. to 7.8 in. (a change of 2.8 in.) is caused by a 4.0 lb weight, so $k = \frac{F}{x} = \frac{4.0}{2.8}$. To find the work done stretching from 5.0 in. to 13.3 in., we use the change in length ($x_2 - x_1$) from the unstretched length (5.0 in.) to the final length (13.3 in.), which is 8.3 in. Plugging these values into the work formula gives us the work done as approximately 49 in.-lb, assuming linear elasticity and Hooke's Law apply throughout the stretch.

Solve the Problem. -The Spring of a Spring Balance Is 5.0 In. Long