Question

3.
The distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 6 inches with 80 pounds attached, how far will it stretch with 55 pounds attached? Round to the nearest tenth of an inch.

Answer 1

Given that the distance a spring stretches varies directly with the weight attached, we can use the formula for direct variation:

\[
d = k \cdot w
\]

where \( d \) is the distance the spring stretches, \( w \) is the weight attached, and \( k \) is the constant of proportionality.

From the problem, we know that when \( w = 80 \) pounds, the spring stretches \( d = 6 \) inches. We can solve for \( k \):

\[
6 = k \cdot 80
\]

Dividing both sides by 80:

\[
k = \frac{6}{80} = \frac{3}{40}
\]

Now that we have \( k \), we can find out how far the spring will stretch with \( w = 55 \) pounds:

\[
d = k \cdot w = \frac{3}{40} \cdot 55
\]

Calculating this:

\[
d = \frac{3 \cdot 55}{40} = \frac{165}{40}
\]

Now, perform the division:

\[
\frac{165}{40} = 4.125
\]

Rounding to the nearest tenth of an inch, we get:

\[
d \approx 4.1 \text{ inches}
\]

Thus, the spring will stretch approximately **4.1 inches** with 55 pounds attached.