A spring with a force constant of 111 N/m is attached to the ceiling. When an unknown mass is attached it pulls the spring down .340 meters to its equilibrium position. The mass is then pulled 9cm below its equilibrium position and oscillates back and forth. Determine the mass (in kg):
To find the mass, we can use Hooke's law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
When the spring is pulled down to its equilibrium position, the force it exerts is given by:
F = k * x
where F is the force, k is the force constant, and x is the displacement from the equilibrium position.
In this case, x = -0.340 m (negative because it is pulled down) and k = 111 N/m.
So, the force exerted by the spring is:
F = 111 N/m * (-0.340 m)
= -37.74 N
Since the force is equal to the weight of the mass hanging from the spring, we can set up the equation:
F = mg
where m is the mass and g is the acceleration due to gravity.
So, -37.74 N = m * 9.8 m/s^2
Solving for m:
m = -37.74 N / 9.8 m/s^2
= -3.85 kg
The mass must be a positive value, so we can take the absolute value:
m = 3.85 kg
Therefore, the mass hanging from the spring is 3.85 kg.