a mass of 2.1 kg attached to a verticalspring stretches the spring 6.2 cm from its original equilibriumposition, what is the spring constant?
Weight in Newtons = m g = 2.1 * 9.81 = F
stretch in meters = 0.062 = x
F = k x
2.1 * 9.81 = k * 0.062
To find the spring constant, we first need to understand Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
First, let's convert the displacement from centimeters to meters, as the standard SI unit for displacement is meters. 6.2 cm is equal to 0.062 meters.
According to Hooke's Law, the force exerted by the spring can be calculated using the formula:
F = -k * x
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 force exerted by the spring when it is stretched 0.062 meters is equal to:
F = -k * 0.062
Now, we need to consider Newton's second law, which states that force equals mass multiplied by acceleration:
F = m * a
Here, F is the force, m is the mass, and a is the acceleration.
Since the spring is being stretched, we can assume the acceleration is zero, as the mass is not moving up or down vertically. Therefore, the force on the mass is balanced by the force exerted by the spring.
F = m * a
F = -k * x
m * a = -k * x
Now, we can rearrange the equation to solve for the spring constant:
k = -m * a / x
In this case, the mass of the object is given as 2.1 kg, and the displacement is 0.062 meters.
k = (-2.1 kg) * (9.8 m/s^2) / (0.062 m)
Calculating this, we find:
k = -33.9 N/m
However, in this case, the negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. Therefore, we can consider the magnitude of the spring constant, resulting in:
k = 33.9 N/m
So, the spring constant is 33.9 N/m.