Question

9.
The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inches with 75 pounds attached, how far will it stretch with 65 pounds attached? Round to the nearest tenth of an inch.

Answer 1

The relationship between the distance a spring stretches and the weight attached can be represented using the concept of direct variation. This means we can set up a proportion based on the information given.

Let \( d \) be the distance the spring stretches, and \( w \) be the weight attached. We know from the problem that:

\[
d_1 = 11 \text{ inches} \text{ when } w_1 = 75 \text{ pounds}
\]

We want to find \( d_2 \) when \( w_2 = 65 \text{ pounds} \). Since the stretching distance varies directly with weight, we can set up the following proportion:

\[
\frac{d_1}{w_1} = \frac{d_2}{w_2}
\]

Substituting in the known values:

\[
\frac{11}{75} = \frac{d_2}{65}
\]

To find \( d_2 \), we can cross-multiply:

\[
11 \times 65 = 75 \times d_2
\]

Calculating \( 11 \times 65 \):

\[
11 \times 65 = 715
\]

So we have:

\[
715 = 75 \times d_2
\]

Now, we solve for \( d_2 \):

\[
d_2 = \frac{715}{75}
\]

Calculating the division:

\[
d_2 \approx 9.4667
\]

Rounding to the nearest tenth of an inch:

\[
d_2 \approx 9.5
\]

Thus, the spring will stretch approximately **9.5 inches** with 65 pounds attached.