A 2.15 kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 18.7 N is required to hold the mass at rest when it is pulled 0.205 m from its equilibrium position (the origin of the x axis). The mass is now released from rest with an initial displacement of xi = 0.205 m, and it subsequently undergoes simple harmonic oscillations. Calculate the force constant of the spring.
18.4N=k(.205) solve for k.
To calculate the force constant of the spring, 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.
Hooke's Law can be expressed as:
F = -kx
Where F is the force exerted by the spring, k is the force constant, and x is the displacement from the equilibrium position.
In this case, we know the force required to hold the mass at rest and the displacement from the equilibrium position. We can use these values to calculate the force constant.
Step 1: Identify the given values
- Mass (m) = 2.15 kg
- Force required to hold the mass at rest (F) = 18.7 N
- Displacement from equilibrium position (x) = 0.205 m
Step 2: Substitute the given values into Hooke's Law
18.7 N = -k * 0.205 m
Step 3: Solve for the force constant (k)
Divide both sides of the equation by -0.205 m:
-18.7 N / 0.205 m = k
k = -91 N/m
Therefore, the force constant of the spring is 91 N/m.