To find out how far the springs will compress with 700.0 kg of cargo, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. This can be represented as:
\[ F = k \cdot x \]
Where:
- \( F \) is the force applied to the spring,
- \( k \) is the spring constant,
- \( x \) is the displacement (compression or extension) of the spring.
When the 600.0 kg crate compresses the springs by 6.0 cm (0.06 m), we can first calculate the force due to the weight of the 600.0 kg crate:
\[ F = m \cdot g \]
Where:
- \( m = 600.0 \) kg (mass of the crate),
- \( g = 9.81 \) m/s² (acceleration due to gravity).
Calculating the force:
\[ F = 600.0 , \text{kg} \times 9.81 , \text{m/s}^2 = 5886 , \text{N} \]
Now we know that a force of 5886 N causes a compression of 0.06 m. We can use this information to find the spring constant \( k \):
\[ k = \frac{F}{x} = \frac{5886 , \text{N}}{0.06 , \text{m}} = 98100 , \text{N/m} \]
Now, we need to find out how much the springs will compress under the weight of 700.0 kg:
\[ F = 700.0 , \text{kg} \times 9.81 , \text{m/s}^2 = 6867 , \text{N} \]
Using the spring constant \( k \) we calculated, we can find the new compression \( x' \):
\[ x' = \frac{F}{k} = \frac{6867 , \text{N}}{98100 , \text{N/m}} \approx 0.07 , \text{m} \]
Thus, the springs will compress approximately 0.07 m when 700.0 kg of cargo is placed in the truck bed.
Therefore, the correct answer is 0.07 m.
1 answer