A spring has a k-value of 1500 N/cm. A 30.0 g object is attached to the spring. Find the force on the object when the spring is compressed 0.01 m beyond equilibrium. Find the force when the spring is stretched 100 cm beyond the equilibrium.
To find the force on the object when the spring is either compressed or stretched, we can use Hooke's Law, which states that the force applied on a spring is directly proportional to the amount it is stretched or compressed.
Hooke's Law equation is given by:
F = -kx
where:
F is the force applied on the spring,
k is the spring constant, and
x is the displacement from the equilibrium position.
To find the force when the spring is compressed by 0.01 m beyond equilibrium:
Step 1: Convert the spring constant from N/cm to N/m.
Since 1 m = 100 cm, the spring constant (k) can be converted to N/m by dividing it by 100.
k = 1500 N/cm / 100 cm/m = 15 N/m
Step 2: Substitute the values into Hooke's Law equation.
F = -kx
F = -(15 N/m)(0.01 m)
F = -0.15 N
The force on the object when the spring is compressed 0.01 m beyond equilibrium is -0.15 N. The negative sign indicates that the force is acting in the opposite direction of the compression.
To find the force when the spring is stretched 100 cm beyond equilibrium:
Step 1: Convert 100 cm to meters.
100 cm = 1 m
Step 2: Substitute the values into Hooke's Law equation.
F = -kx
F = -(15 N/m)(1 m)
F = -15 N
The force on the object when the spring is stretched 100 cm beyond equilibrium is -15 N. Again, the negative sign indicates that the force is acting in the opposite direction of the stretch.