If a mass of 65kg drops 4.3m onto a spring with a stiffness coefficient of 6.2x10n/m to the 4th power, how far does it compess the spring?
To find out how far the spring compresses, 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 equation is given by:
F = -kx
Where:
F is the force exerted by the spring (in newtons),
k is the stiffness coefficient (in N/m^4),
x is the displacement of the spring (in meters).
In this case, we need to solve for x (displacement). The first step is to calculate the force exerted by the spring.
Using the equation:
F = mg
Where:
m is the mass (in kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (65 kg) * (9.8 m/s^2)
F ≈ 637 N
Now, we can substitute the force and the stiffness coefficient into Hooke's Law and solve for x:
637 N = -(6.2 x 10^4 N/m^4) * x
To isolate x, we divide both sides by -(6.2 x 10^4 N/m^4):
x = 637 N / (6.2 x 10^4 N/m^4)
x ≈ 0.0103 m
Therefore, the spring compresses approximately 0.0103 meters (or 10.3 millimeters) when a mass of 65 kg drops 4.3 meters onto it.