Question

Hooke's Law In Exercises 3-10, use Hooke's Law to determine the variable force in the spring problem. An overhead garage door has two springs, one on each side of the door. A force of 15 pounds is required to stretch each spring I foot. Because of the pulley system, the springs stretch only one-half the distance the door travels. The door moves a total of 8 feet and the springs are at their natural length when the door is open. Find the work done by the pair of springs.

Short answer

The work done by the pair of springs when the door moves 8 feet is 240 foot-pounds.

Step-by-step solution

Determine the Spring Constant

From the problem, it is understood that a force of 15 pounds stretches each spring by 1 foot. Therefore, the spring constant \(k\) is 15 lbs/ft.

Calculate the Distance Each Spring Stretches

The door moves a total of 8 feet and because of the pulley system, the springs stretch only half the distance. Therefore, the distance each spring stretches (\(x\)) is 8 / 2 = 4 feet.

Calculate the Work Done by One Spring

Work done to stretch or compress a spring is given by \(W = \frac{1}{2}kx^2\). Substituting the values for \(k\) (15 lbs/ft) and \(x\) (4 ft), we get \(W = \frac{1}{2} * 15 * (4^2) = 120\) foot-pounds.

Compute the Total Work Done by Both Springs

Since there are two springs and they both do the same amount of work, the total work done is twice the work done by one spring. Therefore, the total work is \(2 * W = 2 * 120 = 240\) foot-pounds.