A mass of 0.52 kg attached to a vertical spring
stretches the spring 39 cm from its original
equilibrium position.
The acceleration of gravity is 9.81 m/s2 .
What is the spring constant?
force=kx
.52g=k(.39)
solve for k
13.1 N/m
To find the spring constant, 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 (in newtons)
k is the spring constant (in newtons per meter)
x is the displacement from the equilibrium position (in meters)
In this case, the mass is pulling the spring downwards, so the force exerted by the spring is equal to the weight of the mass, given by: F = mg
Now, we can set up the equation by equating the force exerted by the spring to the weight of the mass:
mg = -kx
Using the known values:
m = 0.52 kg (mass)
g = 9.81 m/s^2 (acceleration due to gravity)
x = 0.39 m (displacement)
Substituting these values into the equation:
(0.52 kg)(9.81 m/s^2) = -k(0.39 m)
Simplifying the equation:
5.1096 N = -0.39 k
Dividing both sides by -0.39:
k = 5.1096 N / 0.39 m
Calculating the spring constant:
k = 13.13 N/m
Therefore, the spring constant is approximately 13.13 N/m.