To find the acceleration of the two objects and the tension in the cord, we need to analyze the forces acting on both the ball and the block.
Forces acting on the block (m2 = 5 kg):
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Weight (mg): The force due to gravity acting on the block is \( F_g = m_2 g \) where \( g \approx 9.81 , \text{m/s}^2 \). \[ F_g = 5 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 49.05 , \text{N} \]
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Normal force (N): The normal force acts perpendicular to the incline and is given by \( N = m_2 g \cos(\theta) \). \[ N = 5 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot \cos(30^\circ) = 5 \cdot 9.81 \cdot \frac{\sqrt{3}}{2} \approx 42.44 , \text{N} \]
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Frictional force (F_f): The kinetic friction force \( F_f = \mu_k N \). \[ F_f = 0.3 \cdot 42.44 \approx 12.73 , \text{N} \]
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Component of gravitational force parallel to the incline (F_g_parallel): \[ F_g\text{_parallel} = m_2 g \sin(\theta) = 5 \cdot 9.81 \cdot \sin(30^\circ) = 5 \cdot 9.81 \cdot 0.5 \approx 24.53 , \text{N} \]
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Net force acting on the block: The net force \( F_{net} \) acting on the block down the incline is: \[ F_{net} = F_g\text{_parallel} - F_f - T \] where \( T \) is the tension in the cord.
Forces acting on the ball (m1 = 1 kg):
The only force acting on the ball is its weight \( F_{g1} = m_1 g \), which acts downward: \[ F_{g1} = 1 \cdot 9.81 = 9.81 , \text{N} \]
Setting up the equations:
Using Newton's second law \( F = ma \):
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For the block (m2): \[ F_{net} = m_2 a \Rightarrow m_2 g \sin(30^\circ) - F_f - T = m_2 a \] Thus: \[ 24.53 - 12.73 - T = 5a \quad \text{(1)} \]
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For the ball (m1): \[ m_1 g - T = m_1 a \] Thus: \[ 9.81 - T = 1a \quad \text{(2)} \]
Solving the equations:
From Equation (2): \[ T = 9.81 - a \quad \text{(3)} \] Substituting Equation (3) into Equation (1): \[ 24.53 - 12.73 - (9.81 - a) = 5a \] This simplifies to: \[ 24.53 - 12.73 - 9.81 + a = 5a \] \[ 2.99 + a = 5a \] \[ 2.99 = 4a \] \[ a = \frac{2.99}{4} \approx 0.7475 , \text{m/s}^2 \]
Now substituting back to find the tension \( T \) using Equation (3): \[ T = 9.81 - 0.7475 \approx 9.0625 , \text{N} \]
Final Results:
- The magnitude of the acceleration of the two objects is approximately \( 0.75 , \text{m/s}^2 \).
- The tension in the lightweight cord is approximately \( 9.06 , \text{N} \).