Questions: Two objects are pulled vertically upward against gravity as shown. Determine the acceleration of the objects for m₁=6.0 kg, m₂=4.0 kg and F=150 N.

Solution Steps

For each object, we need to consider the gravitational force and the applied force. The gravitational force acting on each object is given by \( F_{\text{gravity}} = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).

For \( m_1 = 6.0 \, \text{kg} \): \[ F_{\text{gravity,1}} = 6.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 58.8 \, \text{N} \]

For \( m_2 = 4.0 \, \text{kg} \): \[ F_{\text{gravity,2}} = 4.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \]

The total gravitational force acting on the system is the sum of the gravitational forces on both objects: \[ F_{\text{gravity,total}} = F_{\text{gravity,1}} + F_{\text{gravity,2}} = 58.8 \, \text{N} + 39.2 \, \text{N} = 98.0 \, \text{N} \]

The net force \( F_{\text{net}} \) acting on the system is the applied force minus the total gravitational force: \[ F_{\text{net}} = F - F_{\text{gravity,total}} = 150 \, \text{N} - 98.0 \, \text{N} = 52.0 \, \text{N} \]

The total mass of the system is the sum of the masses of both objects: \[ m_{\text{total}} = m_1 + m_2 = 6.0 \, \text{kg} + 4.0 \, \text{kg} = 10.0 \, \text{kg} \]

Using Newton's second law, \( F = m \cdot a \), we can solve for the acceleration \( a \): \[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{52.0 \, \text{N}}{10.0 \, \text{kg}} = 5.2 \, \text{m/s}^2 \]

Final Answer