To solve these problems, we can use Newton's second law of motion, which states that \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration of the object.
(a) Wayne’s acceleration
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Given Data:
- Force applied by Ancilla on Wayne, \( F = 200 , \text{N} \) (directed South)
- Mass of Wayne, \( m_W = 83 , \text{kg} \)
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Using Newton's second law: \[ a_W = \frac{F}{m_W} \]
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Substituting the known values: \[ a_W = \frac{200 , \text{N}}{83 , \text{kg}} \approx 2.41 , \text{m/s}^2 , [S] \]
Thus, Wayne’s acceleration is approximately \( 2.41 , \text{m/s}^2 \) directed South.
(b) Force exerted by Wayne on Ancilla
According to Newton's third law, for every action, there is an equal and opposite reaction. This means that the force exerted by Wayne on Ancilla will be equal in magnitude but opposite in direction to the force exerted by Ancilla on Wayne.
- Force exerted by Wayne on Ancilla: \[ F = -200 , \text{N} , [N] \] (Here, the negative sign indicates that the force direction is opposite to that of the push Wayne received, meaning it is directed North.)
So, Wayne exerts a force of \( 200 , \text{N} \) [N].
(c) Ancilla’s acceleration
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Given Data:
- Force exerted by Wayne on Ancilla, \( F = 200 , \text{N} \) (directed North)
- Mass of Ancilla, \( m_A = 57 , \text{kg} \)
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Using Newton's second law: \[ a_A = \frac{F}{m_A} \]
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Substituting the known values: \[ a_A = \frac{200 , \text{N}}{57 , \text{kg}} \approx 3.51 , \text{m/s}^2 , [N] \]
Thus, Ancilla’s acceleration is approximately \( 3.51 , \text{m/s}^2 \) directed North.
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