If the trampoline can be considered as a spring with a spring constant of 4.5 x 10^5 N/m, what is the maximum depression of the trampoline?
I started with this knowledge:
E=1/2mv^2+mgy+1/2kx^2
This being the total of energies.
I have done a similar problem in an example problem but I do not know how to do this one.
So here is the work I have so far:
a= highest point
b=hit trampoline
c=lowest point
Ea= 1/2(75)(v^2)+(75)(9.80)(3.2)+1/2k(0)^2
=37.5 v^2 +2352
Eb= 1/2(75)(v^2)+(75)(9.80)(0)+1/2k(0)^2
=37.5 v^2
Ec= 1/2(75)(0)^2+(75)(9.80)(x)+1/2 (4.5 x 10^5)x^2
=735x+2.25 x10^5 x^2
I am unsure of where to go from here.
To find the maximum depression of the trampoline, we need to determine the position where the gravitational potential energy is converted entirely into the potential energy stored in the spring. This occurs when the kinetic energy of the person is zero.
We can start by setting Ea (the total energy at the highest point) equal to Ec (the total energy at the lowest point), since the potential energy is converted to spring potential energy:
37.5v^2 + 2352 = 735x + 2.25 x 10^5 x^2
Now, let's look for the maximum depression (maximum value of x) by taking the derivative of the equation with respect to x and setting it equal to zero:
d/dx (37.5v^2 + 2352) = d/dx (735x + 2.25 x 10^5 x^2)
0 = 735 + 4.5 x 10^5 x
Now, solve for x:
4.5 x 10^5 x = -735
x = -735 / (4.5 x 10^5)
x ≈ -0.00163 meters
However, negative values do not make sense in this context since we are looking for a depression. So, we discard the negative solution.
Therefore, the maximum depression of the trampoline is approximately 0.00163 meters.