When a .90 kg object is attached to a vertically supported spring, it stretches .15 m from the equilibrium position. Find k for the spring.
To find the spring constant, k, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it 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 from the equilibrium position.
In this case, the object is attached vertically, so we need to consider the gravitational force as well. When the object is at rest, the gravitational force is balanced by the force exerted by the spring. Therefore, we can set up the equation:
mg = kx
Where:
m is the mass of the object (0.90 kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
x is the displacement from the equilibrium position (0.15 m).
To find k, we can rearrange the equation:
k = mg / x
Substituting the given values, we have:
k = (0.90 kg) * (9.8 m/s^2) / (0.15 m)
Performing the calculation yields:
k ≈ 58.8 N/m
Therefore, the spring constant, k, is approximately 58.8 N/m.