Question

*Look at attachment for photo of object**

An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).

The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.

(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N

(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2


(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.

Answer 1

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).

A. How to determine the force

• Extension (e) = 0.150 m
• Spring constant (K) = 106 N/m
• Force (F) = ?

F = Ke

F = 106 × 0.15

F = 15.9 N

B. How to determine the acceleration

• 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

C. How to determine the direction of the acceleration vector

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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