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- 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 :/

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 :/