A 50.0 kg girl bounces up and down on a pogo stick. The girl has an instantaneous acceleration of 2.0 m/s^2 when the displacement is -8.0cm. What is the spring constant of the pogo sticks spring?
The spring force at that time (if accelerating upwards) is
F = m (g + a) = 590 N
The spring constant is 590 N/0.08 m
To find the spring constant of the pogo stick's spring, 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. The formula for Hooke's Law is:
F = -kx
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, we are given the mass of the girl (m = 50.0 kg), the acceleration (a = 2.0 m/s^2), and the displacement (x = -8.0 cm).
We can relate the acceleration to the force using Newton's second law:
F = ma
Substituting the given values, we have:
ma = -kx
Rearranging the equation, we get:
k = -(ma) / x
Now we can plug in the given values:
k = -((50.0 kg)(2.0 m/s^2)) / (-0.08 m)
Simplifying the equation further, we have:
k = 125 N/m
Therefore, the spring constant of the pogo stick's spring is 125 N/m.