Question

A spring with spring constant 0.5 cmN has a force of 40 newtons applied to it. How much does the spring stretch? Use Hooke's law F = kx. (1 point) Responses 1,600 cm 1,600 cm 100 cm 100 cm 80 cm 80 cm 40 cm 40 cm

Answer 1

To find out how much the spring stretches, we can use Hooke's law, which is given by the formula:

\[ F = kx \]

where:
- \( F \) is the force applied to the spring (in newtons),
- \( k \) is the spring constant (in N/cm), and
- \( x \) is the stretch of the spring (in cm).

Given:
- \( F = 40 \) N
- \( k = 0.5 \) cmN

We first need to convert the spring constant \( k \) to N/m (since there are 100 cm in a meter):

\[
k = 0.5 \, \text{cmN} = 0.5 \times 0.01 \, \text{mN} = 0.0005 \, \text{N/m}
\]

Now we can rearrange Hooke's law to solve for \( x \):

\[
x = \frac{F}{k}
\]

Substituting the values:

\[
x = \frac{40 \, \text{N}}{0.0005 \, \text{N/m}} = 80000 \, \text{m}
\]

Since the options provided are significantly smaller, let’s double-check the units. The initial value of \( k \) provided in the question as 0.5 cmN is likely intended as 0.5 N/m, which would apply directly without conversion for the context.

Revisiting with:

\( k = 0.5 \) N/m:
\[
x = \frac{40 \, \text{N}}{0.5 \, \text{N/m}} = 80 \, \text{m}
\]

In cm, that is:
\[
80 \, \text{m} = 8000 \, \text{cm}
\]

Therefore, the spring stretches by **80 cm**.

Hence, the correct answer is:
**80 cm**.