When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x?

When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x?

To determine the amount of stretch, x, when a mass m=0.50kg is attached to the springs, we need to know the spring constant, k, of the springs.

The amount of stretch of a spring when a mass is attached to it is determined by Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring. Mathematically, Hooke's Law can be expressed as:

F = -kx

Where F is the force exerted by the spring, k is the spring constant, and x is the displacement or stretch of the spring.

To find x, we can rearrange Hooke's Law:

x = -F/k

Assuming that the force exerted by the spring is equal to the weight of the mass (F = mg), we can substitute the values into the equation:

x = -(mg)/k

Using the given values, m = 0.50kg and g = acceleration due to gravity (approximately 9.8 m/s^2), we still need the value of k to calculate x accurately.

To determine the amount of stretch, x, when a mass m = 0.50kg is attached to the springs, we need to use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring from its equilibrium position.

The equation for Hooke's Law is given as:
F = kx

Where:
F = force exerted by the spring
k = spring constant
x = displacement or stretch of the spring

To find the amount of stretch, x, we need to know the spring constant, which is a property of the specific spring used in the setup.

Please provide the value of the spring constant, k, in order to calculate the amount of stretch, x.

depends on the spring constant.

asking twice won't help ...