Answer: approximately -170.43 N/m.
Step-by-step explanation:
To analyze the situation, 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, this can be expressed as:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant,
x is the displacement from the equilibrium position.
Given that a mass of 2.00 kg is hung from the spring and causes it to stretch 11.5 cm (or 0.115 meters), we can find the force exerted by the spring using the formula:
F = mg
Where:
m is the mass, which is 2.00 kg,
g is the acceleration due to gravity, approximately 9.8 m/s².
F = 2.00 kg * 9.8 m/s²
F = 19.6 N
Now, we can find the spring constant (k) using the equation:
F = -kx
k = -F / x
k = -19.6 N / 0.115 m
k ≈ -170.43 N/m
The negative sign indicates that the force exerted by the spring opposes the displacement, which is a characteristic of a restoring force.
Therefore, the spring constant for the given situation is approximately -170.43 N/m.