By how much will a 2.05 kg object stretch a spring that is suspended vertically if the spring constant is 126 N/m?

M*g = 2.05 * 9.8 = 20.1 N. = force of the object.

d = 1m/126N * 20.1N. =

To determine the amount of stretch in a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

The formula for Hooke's Law is:

F = k * x

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

In this case, we are given the mass of the object, which we can use to calculate the force exerted by gravity. We can then equate this force with the force exerted by the spring to find the displacement.

Step 1: Calculate the force exerted by gravity.
The force exerted by gravity can be calculated using Newton's second law:

F = m * g

Where:
F is the force exerted by gravity,
m is the mass of the object, and
g is the acceleration due to gravity (approximated as 9.8 m/s^2).

In this case, the mass is given as 2.05 kg, so:

F = 2.05 kg * 9.8 m/s^2.

Step 2: Equate the force exerted by gravity with the force exerted by the spring.
From Hooke's Law, we know that:

F = k * x

Therefore:

2.05 kg * 9.8 m/s^2 = 126 N/m * x

Step 3: Solve for x.
Rearranging the equation, we find:

x = (2.05 kg * 9.8 m/s^2) / 126 N/m

Calculating the values:

x = 20.09 m^2/s^2 / 126 N/m

x = 0.1597 m

Therefore, the 2.05 kg object will stretch the spring by approximately 0.1597 meters.