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

[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 :/

 

[malawi_glenn]

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.

 

[Steve4Physics]

$$\frac{1}{2}mv_0^2 = mgh_1 + \frac{1}{2}kx_1^2$$

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.