A 1.0 kg object is suspended from a vertical spring whose spring constant is 105 N/m. The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.

The original position is the equilibrium position, and the velocity is a maximum there. Let that velocity be Vmax

(1/2) M Vmax^2 = (1/2) k X^2

Vmax = X*sqrt(k/M)

k is the spring constant
X = 0.25 m
m = 1.0 kg

I have not considered gravitational potential energy in this derivation. As I recall, its effect cancels out when you define X as being measured from the equilibrium position, with stretching due to the weight taken into account. I am too lazy to verify this.

To find the speed with which the object passes through its original position on the way up, we can use the principles of energy conservation.

First, let's find the potential energy of the spring when the object is at its original position. The potential energy stored in a spring is given by the formula:

Potential Energy = (1/2) * k * x^2

Where k is the spring constant and x is the displacement from the equilibrium position.

Given that the spring constant (k) is 105 N/m and the object is pulled down by an additional distance of 0.25 m, the total displacement (x) from the original position is 0.25 m.

Potential Energy at the original position = (1/2) * 105 N/m * (0.25 m)^2
Potential Energy at the original position = 3.28125 J

According to the law of conservation of energy, the potential energy at the original position will be converted into kinetic energy when the object passes through this point.

Kinetic Energy = Potential Energy at the original position

So, we can find the kinetic energy of the object at this point.

Kinetic Energy = 3.28125 J

Kinetic Energy is given by the formula:

Kinetic Energy = (1/2) * m * v^2

Where m is the mass of the object and v is the velocity.

Given that the mass (m) is 1.0 kg, we can solve for the velocity (v).

Kinetic Energy = (1/2) * 1.0 kg * v^2
3.28125 J = 0.5 kg * v^2

Dividing both sides of the equation by 0.5 kg:

6.5625 J/kg = v^2

Taking the square root of both sides of the equation:

v = √(6.5625 J/kg)

Calculating the square root:

v ≈ 2.56 m/s

Therefore, the speed with which the object passes through its original position on the way up is approximately 2.56 m/s.