A cube, whose mass is 0.660 kg, is attached to a spring with a force constant of 114 N/m. The cube rests upon a frictionless, horizontal surface (shown in the figure below).
A cube labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the cube is displaced by a distance A to the right of its equilibrium position.
The cube is pulled to the right a distance A = 0.150 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?
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).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.
Correct: Your answer is correct.
To find the answers to these questions, we'll need to use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's law is:
F = -kx
Where F is the force exerted by the spring, k is the force constant of the spring, and x is the displacement from the equilibrium position.
(a) To find the magnitude of the spring force acting on the cube at the instant of release, we can use Hooke's law. We're given the force constant k, which is 114 N/m, and the displacement A, which is 0.150 m.
F = -kx
F = -(114 N/m)(0.150 m)
F = -17.1 N
The magnitude of the spring force acting on the cube at the instant of release is 17.1 N.
(b) To find the magnitude of the cube's acceleration at that instant, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
F_net = m * a
In this case, the net force is the spring force, which we just calculated to be 17.1 N. The mass of the cube is given as 0.660 kg.
17.1 N = (0.660 kg) * a
Solving for a gives us:
a = 17.1 N / 0.660 kg
a ≈ 25.91 m/s^2
The magnitude of the cube's acceleration at that instant is approximately 25.91 m/s^2.
(c) The direction of the acceleration vector can be determined by the direction of the net force acting on the cube. Since the spring force is negative (due to the negative sign in Hooke's law), it means that the force is acting in the opposite direction of the displacement. Therefore, the acceleration vector points towards the equilibrium position (to the left in the figure).
So, the correct answer is: Toward the equilibrium position (i.e., to the left in the figure).