To solve the problem, we can use Newton's second law of motion, which states that \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
(a) Wayne's acceleration
Given:
- Force (F) exerted by Ancilla on Wayne: 200 N (in the South direction)
- Mass of Wayne (m): 83 kg
We can calculate Wayne's acceleration (a) using the formula:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{200 , \text{N}}{83 , \text{kg}} \approx 2.41 , \text{m/s}^2 \]
So, Wayne's acceleration is approximately 2.41 m/s² [S].
(b) Force that Wayne exerts 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 is equal in magnitude but opposite in direction to the force exerted by Ancilla on Wayne.
Therefore, the force that Wayne exerts on Ancilla is 200 N [N] (in the North direction).
(c) Ancilla’s acceleration
Using the same approach for Ancilla:
Given:
- The magnitude of the force that Wayne exerts on Ancilla: 200 N (upwards, or North)
- Mass of Ancilla (m): 57 kg
Calculating Ancilla's acceleration (a):
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{200 , \text{N}}{57 , \text{kg}} \approx 3.51 , \text{m/s}^2 \]
So, Ancilla's acceleration is approximately 3.51 m/s² [N] (or North).
Summary:
- (a) Wayne's acceleration: 2.41 m/s² [S]
- (b) Force exerted by Wayne on Ancilla: 200 N [N]
- (c) Ancilla's acceleration: 3.51 m/s² [N]
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