Homework 6 section 3.5 (Homework)

Question

Hooke's law states that the distance a spring stretches varies directly as the weight on the spring. Let d represent the distance (in inches) the spring stretches, let w represent the weight (in pounds) on the spring, and let k be the constant of proportionality. Write an equation that relates d to w.

d = 

 

A weight of 50 pounds stretches a spring 4 inches.

4 in.

50 lb

Two springs hang vertically; each connects at one end to a horizontal ceiling.

1. The first spring hangs freely from the left end of the ceiling.
2. The second spring has a weight attached to its free end, extends down further than the first spring, and hangs from the right end of the ceiling. The weight is labeled 50 lb.
3. A dashed horizontal line extends from the bottom of the second spring (just above the weight) to the left.
4. A vertical arrow extends from the dashed horizontal line up to the bottom of the first spring. The vertical arrow is labeled 4 in.

Use the equation that relates d to w and the known information to find k.

k = 

How far (in inches) will a weight of 160 pounds stretch the spring?

Answer 1

If the distance d a spring stretches is directly proportional to the weight w attached to the spring, then we can write an equation as follows:

d = kw . . . . . . (1)

solving for k we get the relationship:

k = d / w . . . . . . (2)

In the first instance of a 50 Ib weight, which stretches the spring by 4 inches, we can solve for the constant k using the relationship in 2) above

Therefore k = 4 inches / 50 ibs

= 2 inches / 25 Ibs

Slotting this value of k into (1) above, we can restate the equation as

d = (2 inches / 25 Ibs) * w . . . . . . . (3)

To find the distance that a spring stretches with a 160 Ib weight on it, we use equation (3) above as follows:

d = (2 inches / 25 Ibs) * 160 Ibs

canceling out the Ibs, leaves us with inches, gives us

d = 320 / 25 inches

d = 12.8 inches. This is the distance a 160 Ib weight will stretch the spring to.