A mass of .50 kg attached to a vertical spring stretches the spring 37 cm from its original position. The accel of gravity is 9.81 m/s2. What is the spring constant?? Answer in units of N/m
force=kx
mg=kx
you know, m, g, x solve for k Mass needst to be in kg, of course, and x in meters.
To find the spring constant, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is:
F = k * x
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
In this case, we know the mass of the object (0.50 kg) and the displacement of the spring (37 cm), but we need to convert the displacement to meters since the acceleration due to gravity is given in m/s^2.
First, let's convert the displacement to meters:
37 cm = 0.37 m
Now, we can find the force exerted by the spring using Newton's second law:
F = m * g
where m is the mass (0.50 kg) and g is the acceleration due to gravity (9.81 m/s^2).
F = (0.50 kg) * (9.81 m/s^2)
F = 4.905 N
Now, we can substitute the values into Hooke's Law:
4.905 N = k * 0.37 m
To solve for k, we can rearrange the equation:
k = (4.905 N) / (0.37 m)
k ≈ 13.27 N/m
Therefore, the spring constant is approximately 13.27 N/m.