Physics question, involving work-energy theorem.?

Question

Consider a bungee jumper of mass 65 kg, with a 9.00m cord tied to his ankles. When stretched this cord may be treated as a spring, of spring constant 150N/m. Note that the jump off point is at 9.00m and the origin for the x-coordinate is at the point where the rubber cord becomes taut.
a.)At what point does the jumper attain maximum speed(I got -4.25 meters)
b)What is the value of this maximum speed?
c)At what point does the jumper attain maximum acceleration?
d)What is the value of this maximum acceleration?

**I got part a by using E=mgx+1/2(kx^2) and then taking the derivative since de/dx is 0 and then solving for x. Answer and explanation as to why you did what you did would be very helpful as I am very confused about this problem.

NCS · Accepted Answer

Maybe you're right about part (a). I would do it differently. The velocity increases until the net acceleration is zero, and that happens when the spring force and the weight sum to zero.
mg + kx = 0
x = -mg / k = 65kg * 9.8m/s² / 150N/m = -4.25 m ◄ Sweet.

b) At top, E = PE = mgh = 65kg * 9.8m/s² * 9m = 5733 J

At max velocity, E = PE + U + KE
5733 J = 65kg * 9.8m/s² * -4.25m + ½ * 150N/m * (4.25m)² + ½ * 65kg * v²
5733 J = -1353 J + 32.5kg * v²
v² = 7086J / 32.5kg = 218 m²/s²
v = 14.8 m/s ◄

c) Maximum acceleration is either during free-fall (-9.8 m/s²) or at maximum stretch (more likely). At that point, the KE is zero, and
E = PE + U
5733 J = 65kg * 9.8m/s² * x + ½ * 150N/m * x²
which is quadratic with two roots: x = 5.47 m ← not possible; x must be < 0
and x = -14.0 m. ◄ I'll take this one.

d) acceleration a = g - kx/m = -9.8m/s² - 150N/m*-14m / 65kg = 22.5 m/s² ◄
Oh yeah, that's a maximum.

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Physics question, involving work-energy theorem.?

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