Question

A cube, whose mass is 0.660 kg, is attached to a spring with a force constant of 106 N/m. The cube rests upon a frictionless, horizontal surface (shown in the figure below).

The cube is pulled to the right a distance A = 0.140 m from its equilibrium position (the vertical dashed line) and held motionless. The cube is then released from rest.

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



(b) At that very instant, what is the magnitude of the cube'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).
-Away from the equilibrium position (i.e., to the right in the figure).
- Toward the equilibrium position (i.e., to the left in the figure).
- You cannot tell without more information.


***Please refer to attached photo

Answer 1

F = - F x force acting on cube

F = -106 N/n * .14 m = -14.8 N direction opposite to force

a = F / m = -14.8 N / .66 kg = -22.5 m/s^2 acceleration of cube

direction is towards equilibrium point (opposite F)