Why doesn't this solution work? (Springs and Conservation of Energy)

• CCoffman
In summary, the solution to finding the initial velocity and distance of a trampoliner can be solved using the conservation of energy equation and a kinematic equation. However, using the force method may give a slightly different answer due to the changing force of the trampoline as it is stretched. This difference can be accounted for by including the total height change in the conservation of energy equation. This is not a dumb question and comparing different methods is a good learning strategy.f

CCoffman

Homework Statement
A 67 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7 m/s from a height of 2.0 meters. If the trampoline behaves like a spring of spring constant ##5.8×10^4## N/m , what is the distance he depress it?
Relevant Equations
##U_i + K_i + \text{[Other initial energies] =} U_f + K_f + \text{[Other final energies], }F = kx##
I already know the solution to this, all you do is set the height of the top of the trampoline to 0 and solve for initial velocity so the equation for the conservation of energy $$mgh_0 + \frac{1}{2}mv_0^2 + \frac{1}{2}kx_0^2 = mgh_1 + \frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2$$ becomes $$\frac{1}{2}mv_0^2 = mgh_1 + \frac{1}{2}kx_1^2$$ All we do is solve for the initial velocity using a kinematic equation and then find the distance the spring/trampoline is depressed using the quadratic formula, giving us x = -.28 meters.

On my first attempt i tried to take the simpler rout of using $$F = kx$$ $$ma = kx$$ $$\frac{67*-9.8}{5.8*10^4} = -.011$$ I am wondering why this is incorrect. The acceleration should stay the same, at least the initial acceleration before the energy is transferred to the spring. I am sorry if this is a dumb question, i just started physics and... i don't understand anything :/

I am wondering why this is incorrect.
That will just give you that the spring would compress 1.1 cm when the person is standing still on the trampoline.

$$\frac{1}{2}mv_0^2 = mgh_1 + \frac{1}{2}kx_1^2$$
This should be
##\frac{1}{2}mv_0^2 = mg(h_1+x_1) + \frac{1}{2}kx_1^2##
because the total height-change is ##h_1+ x_1##. But doing this makes only a small difference to the answer if ##x_1 \ll h_1##; also it means you would need to solve a more complicated quadratic equation.

The acceleration should stay the same, at least the initial acceleration before the energy is transferred to the spring.
No. Constant acceleration requires a constant force (F=ma). The acceleration is not constant while the trampoline (spring) is being stretched.

That's because the upwards force from the trampoline increases the more it is stretched (F = -kx and x is changing).

The trampoliner, while in contact with the trampoline, is comparable to a mass hanging on a spring in simple harmonic motion if you have met this before. A bit of calculus is needed for this ‘force method’.

Im sorry if this is a dumb question,
It's not a dumb question. And trying to check the consistency of two different methods to solve the same problem is an excellent learning technique.