Question

A 0.40 kg object is attached to a spring with force constant 159 N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when the spring is compressed 0.12 m.

(a) Find the force on the object.
_____ N
(b) Find its acceleration at that instant
____m/s²

Answer 1

Answer 2

(a) To find the force on the object, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:

F = -kx

Where: F is the force applied by the spring, k is the force constant of the spring, and x is the displacement from the equilibrium position.

In this case, the object is released from rest when the spring is compressed by 0.12 m. Since the object is attached to the spring, the displacement x is equal to the compression of the spring.

Given: k = 159 N/m x = 0.12 m

Substituting these values into the formula, we get:

F = -(159 N/m) * (0.12 m) F = -19.08 N

Therefore, the force on the object is -19.08 N. The negative sign indicates that the force is directed opposite to the displacement.

(b) To find the acceleration of the object at that instant, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force applied to it divided by its mass:

a = F/m

Given: F = -19.08 N m = 0.40 kg

Substituting these values into the formula, we get:

a = (-19.08 N) / (0.40 kg) a = -47.7 m/s^2

Therefore, the acceleration of the object at that instant is -47.7 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the applied force.

. Nonparametric statistics might be preferred under which of the following conditions? Your sample size is very small. Your sample size is very large. The variables you are analyzing are continuous. All the assumptions of parametric statistics have been met.