To find the value of force X in the free-body diagram of the kite, we can analyze the information given.
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The kite is not accelerating vertically, which means that the forces acting in the vertical direction must be equal. The forces in the vertical direction are the weight of the kite and the vertical component of the tension in the kite string.
The weight of the kite can be calculated using the formula:
\( W = m \times g \)
where:
\( m = 2 , \text{kg} \) (mass of the kite)
\( g = 9.81 , \text{m/s}^2 \) (acceleration due to gravity)\( W = 2 , \text{kg} \times 9.81 , \text{m/s}^2 = 19.62 , \text{N} \)
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Since the kite is not accelerating vertically, the vertical component of the tension in the string (let's denote it as \( T_y \)) must equal the weight of the kite, which is approximately 20 N (for simplicity, we round 19.62 N to 20 N).
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The kite is accelerating horizontally to the right at a rate of \( 2 , \text{m/s}^2 \). To find the net force acting in the horizontal direction, we apply Newton's second law:
\( F_{net} = m \times a \)
where:
\( a = 2 , \text{m/s}^2 \)\( F_{net} = 2 , \text{kg} \times 2 , \text{m/s}^2 = 4 , \text{N} \)
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In the horizontal direction, the forces acting on the kite will be the horizontal component of the tension in the string (let's denote it as \( T_x \)) and the force \( X \). Since the kite is accelerating to the right, the net force equation is:
\( T_x - X = F_{net} \)
\( T_x - X = 4 , \text{N} \) -
If we assume that the horizontal component of the tension in the string is what is providing the net force required for the acceleration, and without loss of generality, we can reason that force \( X \) must be the force balancing out this tension.
Since the horizontal net force required is \( 4 , \text{N} \), this force \( X \) also must be 0 N, as any value greater than 0 would indicate that \( T_x \) must be greater than \( 4 , \text{N} \) to account for \( X\).
Based on the analysis, the value of force \( X \) in the diagram is:
B. 0 N
2 answers