To solve the problem, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)).
(a) Wayne's acceleration
Given:
- Force exerted by Ancilla on Wayne, \( F = 200 , \text{N} \) [S]
- Mass of Wayne, \( m_W = 83 , \text{kg} \)
Using Newton's second law: \[ a_W = \frac{F}{m_W} \] Substituting the values: \[ a_W = \frac{200 , \text{N}}{83 , \text{kg}} \approx 2.41 , \text{m/s}^2 , \text{[S]} \]
(b) Force exerted by Wayne on Ancilla
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, the force exerted by Wayne on Ancilla is equal in magnitude but opposite in direction to the force exerted by Ancilla on Wayne.
Thus, the force exerted by Wayne on Ancilla is: \[ F_{WA} = -200 , \text{N} , \text{[N]} \]
(c) Ancilla’s acceleration
Given:
- Mass of Ancilla \( m_A = 57 , \text{kg} \)
- Force exerted by Wayne on Ancilla, \( F = 200 , \text{N} , \text{[N]} \)
Using Newton's second law for Ancilla: \[ a_A = \frac{F}{m_A} \] Substituting the values: \[ a_A = \frac{200 , \text{N}}{57 , \text{kg}} \approx 3.51 , \text{m/s}^2 , \text{[N]} \]
Summary of Results:
- (a) Wayne's acceleration: \( \approx 2.41 , \text{m/s}^2 , \text{[S]} \)
- (b) Force exerted by Wayne on Ancilla: \( -200 , \text{N} , \text{[N]} \)
- (c) Ancilla’s acceleration: \( \approx 3.51 , \text{m/s}^2 , \text{[N]} \)
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