A spring in equilibrium is 10 inches long and an external force of 30 pounds stretches the spring to a length of 18 inches. The work done by the spring in stretching it from equilibrium to 12 inches, in inch-pounds, is

A)-6.5
B)-7.5
C)-8.5
D)-9
E)-9.5

Question

A spring in equilibrium is 10 inches long and an external force of 30 pounds stretches the spring to a length of 18 inches. The work done by the spring in stretching it from equilibrium to 12 inches, in inch-pounds, is
​
A)-6.5
B)-7.5
C)-8.5
D)-9
E)-9.5

Quizplus · Accepted Answer

The work done by a spring can be calculated using the formula $ W = \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the displacement from the equilibrium position. First, we need to find the spring constant $ k $. Given that a force of 30 pounds stretches the spring from 10 inches (its natural length) to 18 inches, the displacement $ x $ is 8 inches. Using Hooke's Law, $ F = kx $, we find $ k = \frac{F}{x} = \frac{30}{8} = 3.75 $ pounds per inch.To find the work done in stretching the spring from 10 inches to 12 inches, the displacement $ x $ is 2 inches. Using the work formula:$$ W = \frac{1}{2} k x^2 = \frac{1}{2} \times 3.75 \times 2^2 = \frac{1}{2} \times 3.75 \times 4 = 7.5 $$ inch-pounds.Since the work is done against the spring force, it is negative, making the correct answer -7.5 inch-pounds.

A Spring in Equilibrium Is 10 Inches Long and an External