The object weighing 20 N is hung on the bottom of the spiral spring , so that stretch as far as 9.8 cm after a state of balance , the weight pulled down as far as 5 cm , measured from idle after burdened . Find the equation of motion generated assuming no friction and air resistance

To find the equation of motion for the given system, we need to first understand some basic concepts of springs and equilibrium.

1. Hooke's Law: It states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as: F = -kx, where F is the force, k is the spring constant, and x is the displacement.

2. Equilibrium Position: It is the position where the net force acting on the object is zero. At this point, the spring force and the gravitational force are balanced.

Now, let's calculate the spring constant (k) using the given information:

1. The weight of the object (force due to gravity) is 20 N.

2. When the system is in equilibrium, the spring is stretched by 9.8 cm (or 0.098 m).

From Hooke's Law, we have:
F = -kx

At equilibrium, the spring force (F) is balanced by the gravitational force (weight):
F = 20 N
x = 0.098 m

Substituting the values, we get:
20 N = -k * 0.098 m

Solving the equation for k, we find:
k = -20 N / 0.098 m ≈ -204.08 N/m

Now, let's find the equation of motion assuming no friction and air resistance:

Let's define the following variables:
m = mass of the object (unknown)
x = displacement of the object from the equilibrium position

Using Newton's second law, we have:
F_net = ma

The net force acting on the object is the sum of the spring force and the gravitational force. Since the spring force is given by Hooke's Law, we can write:
F_spring = -kx

The gravitational force is given by:
F_gravity = mg

Therefore, the net force is:
F_net = F_spring + F_gravity
= -kx + mg

Applying Newton's second law, we have:
ma = -kx + mg

Dividing both sides by m, we get:
a = (-k / m) * x + (g)

This is the equation of motion for the given system, assuming no friction and air resistance. It describes the acceleration (a) of the object as a function of its displacement (x) from the equilibrium position.

Note that the negative sign in front of the spring constant (k) represents the direction of the force exerted by the spring, which opposes the displacement.

To find the equation of motion for the object hanging on the spring, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed.

In this case, we are given that the object weighs 20 N and stretches the spring by 9.8 cm after reaching a state of balance. Additionally, when the weight is pulled down, it is lowered by 5 cm from its idle position.

1. First, let's convert the given measurements into SI units:
- Weight of the object: 20 N
- Stretch after reaching a state of balance: 9.8 cm = 0.098 m
- Lowering when pulled down: 5 cm = 0.05 m

2. Now, let's calculate the spring constant (k) using Hooke's Law:
Hooke's Law states that F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

At the point of balance, the force exerted by the spring (F) is equal to the weight of the object (20 N). So we have:
20 N = -k * 0.098 m

Solving for k, we get:
k = -20 N / 0.098 m = -204.08 N/m

3. Next, let's write the equation of motion:
The equation of motion for an object hanging on a spring is given by:
m * a = -k * x

Where m is the mass of the object and a is the acceleration. Since we are not given the mass, we can cancel it out of the equation. Therefore, the equation of motion becomes:

a = -k * x

Substituting the value of k we found earlier, we have:
a = -204.08 N/m * x

So, the equation of motion for the object assuming no friction and air resistance is:
a = -204.08 * x