To determine the elongation in the spring when the system is in equilibrium, we need to analyze the forces acting on the masses and how they relate to the spring's properties. Given that the spring constant \( k \) is 1000 N/m, and the masses \( m_1 \) and \( m_2 \) are 2 kg and 3 kg respectively, we can start by calculating the gravitational forces acting on each mass.
Calculating Forces on the Masses
The force due to gravity acting on each mass can be calculated using the formula:
- Force on \( m_1 \): \( F_1 = m_1 \cdot g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \)
- Force on \( m_2 \): \( F_2 = m_2 \cdot g = 3 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 30 \, \text{N} \)
Understanding the Equilibrium Condition
In equilibrium, the net force acting on the system must be zero. This means that the upward force exerted by the spring must balance the downward forces due to the weights of the two masses. The spring force can be expressed as:
Spring Force: \( F_s = k \cdot x \)
where \( x \) is the elongation of the spring. Since the system is in equilibrium, we can set up the equation:
Net Force: \( F_s = F_2 - F_1 \)
Substituting the values we calculated:
Net Force: \( k \cdot x = 30 \, \text{N} - 20 \, \text{N} \)
This simplifies to:
Spring Force: \( k \cdot x = 10 \, \text{N} \)
Solving for Elongation
Now we can substitute the spring constant into the equation:
1000 N/m \( \cdot x = 10 \, \text{N} \)
To find \( x \), we rearrange the equation:
Elongation: \( x = \frac{10 \, \text{N}}{1000 \, \text{N/m}} = 0.01 \, \text{m} \)
Final Result
The elongation in the spring when the system is in equilibrium is \( 0.01 \, \text{m} \) or \( 1 \, \text{cm} \). This means that the spring stretches by 1 cm due to the difference in weight between the two masses. Understanding this balance of forces is crucial in mechanics, as it illustrates how different components interact in a system.
