A. The magnitude of the spring force (in N) acting upon the object is 15.9 N
B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²
C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).
<h3>A. How to determine the force </h3>
- Extension (e) = 0.150 m
- Spring constant (K) = 106 N/m
- Force (F) = ?
F = Ke
F = 106 × 0.15
F = 15.9 N
<h3>B. How to determine the acceleration</h3>
- Mass (m) = 0.52 Kg
- Force (F) = 15. 9 N
- Acceleration (a) =?
F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
<h3>C. How to determine the direction of the acceleration vector</h3>
Considering the diagram, we can see that the spring was pulled away from the equilibrium point.
Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.
Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.
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