To solve these problems, we'll use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement (stretch) of the spring from its equilibrium position. The formula is given by:
\[ F = k \cdot x \]
where:
- \( F \) is the force applied to the spring (in Newtons),
- \( k \) is the spring constant (in N/m),
- \( x \) is the displacement of the spring from its rest position (in meters).
a) How far in centimeters would the spring stretch with a mass of 500g?
First, we need to convert the mass from grams to kilograms:
\[ \text{mass} = 500 \text{ g} = 0.5 \text{ kg} \]
Next, we calculate the weight of the object, which is the force due to gravity:
\[ F = m \cdot g = 0.5 \text{ kg} \cdot 10 \text{ N/kg} = 5 \text{ N} \]
Now we can use Hooke's Law to find the stretch \( x \):
\[ F = k \cdot x \]
Rearranging for \( x \):
\[ x = \frac{F}{k} = \frac{5 \text{ N}}{20 \text{ N/m}} = 0.25 \text{ m} \]
To convert \( x \) from meters to centimeters:
\[ x = 0.25 \text{ m} \times 100 \text{ cm/m} = 25 \text{ cm} \]
So, the spring would stretch 25 cm.
b) What is the mass of an object that stretches the spring 35 cm long?
First, convert 35 cm to meters:
\[ x = 35 \text{ cm} = 0.35 \text{ m} \]
Using Hooke's Law again, we can find the force \( F \):
\[ F = k \cdot x = 20 \text{ N/m} \cdot 0.35 \text{ m} = 7 \text{ N} \]
Now, we can find the mass \( m \) that would create this force:
\[ F = m \cdot g \]
Rearranging for \( m \):
\[ m = \frac{F}{g} = \frac{7 \text{ N}}{10 \text{ N/kg}} = 0.7 \text{ kg} \]
So, the mass of the object that stretches the spring 35 cm long is 0.7 kg.