Asked by Sydney
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.
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.
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