To determine the cart's horizontal acceleration, we need to analyze the net force acting on the cart in the horizontal direction, taking into account the two forces applied to it.
- Break down the forces:
- The first force is \( F_1 = 10 , \text{N} \) to the right.
- The second force is \( F_2 = 30 , \text{N} \) at an angle of \( 60^\circ \) above the horizontal, which we will break down into horizontal and vertical components.
The horizontal component \( F_{2,x} \) of the second force can be calculated using the cosine of the angle: \[ F_{2,x} = F_2 \cdot \cos(60^\circ) = 30 , \text{N} \cdot \cos(60^\circ) = 30 , \text{N} \cdot \frac{1}{2} = 15 , \text{N} \]
Since this force is pulling to the left, we take it as negative in the horizontal force equation: \[ F_{2,x} = -15 , \text{N} \]
- Calculate the net horizontal force: Now we can find the net force \( F_{\text{net}} \) acting on the cart in the horizontal direction: \[ F_{\text{net}} = F_1 + F_{2,x} = 10 , \text{N} + (-15 , \text{N}) = 10 , \text{N} - 15 , \text{N} = -5 , \text{N} \]
The negative sign indicates that the net force is directed to the left.
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Calculate the acceleration: Using Newton's second law \( F = ma \), where \( m \) is the mass of the cart: \[ F_{\text{net}} = m \cdot a \] \[ -5 , \text{N} = 10 , \text{kg} \cdot a \] To find the acceleration \( a \): \[ a = \frac{-5 , \text{N}}{10 , \text{kg}} = -0.5 , \text{m/s}^2 \]
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Interpret the result: The negative sign of the acceleration indicates that the acceleration is towards the left.
Final Answer: The magnitude of the cart's horizontal acceleration is \( 0.5 , \text{m/s}^2 \) to the left.