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.