Physics Homework - how to solve?

Question

A 65-kg trampoline artist jumps vertically upward fromt he top of a platform with a speed of 5.0m/s. if hte trampoline behaves like a spring with spring stiffness constant 6.2 x 10^4 N/m, how far does he depress if?

Answer: -0.31 m

Answer 1

I solved this as if the artist strikes the trampoline at 5 m/s downward.
Is he standing on the trampoline and gets sent upward at 5 m/s, or is he standing on a platform some distance above the trampoline?

For striking the trampoline at 5m/s downward
Start by considering the energy in the system:
The artist has kinetic energy of 1/2*m*v^2
and potential energy of m*g*d
where g is 9.8 m/s^2 and d is the net downward displacement of the spring

The spring will absorb the energy =1/2*k*d^2
where k is the spring constant
at equilibrium when the spring is fully compressed the energy in the spring will balance out the rest
set all energy equal to zero for equilibrium
0=1/2*kd^2-m*g*d-1/2*m*v^2
multiply by two
0=k*d^2-2*m*g*d-m*v^2

plug in the numbers
0=6.2E+4*d^2-2*65*9.8*d-65*5^2
0=6.2E+4*d^2-1274d-1625
the positive root, which is the displacement of the spring,
set negative since I used positive motion downward
d=-.172m

Since this isn't the answer, I will assume that the artist was on a platform and jumped onto the trampoline. I will compute the height of the platform above the uncompressed trampoline:
In this case the spring energy is computed from your answer:
1/2(62000)*.31^2
=2979
the potential energy lost due to compression is
65*9.8*.31
=197.5
His vertical velocity when he jumped from the platform
=.5*65*25
=812.5
The energy will sum to zero.
The energy in the spring came from three sources:
His vertical velocity in kinetic energy
Potential energy by standing on a platform above the trampoline
The potential energy given to the spring due to his weight and travel during compression
So
2979-197.5-812.5-65*9.8*h=0
where h is the height of the platform
h=1969/637
h=3.1m

j

Answer 2

The artist depresses the trampoline a distance d during which the kinetic energy is converted to work W in deflecting the spring.

The kinetic energy = the original potential energy from the platform height H plus the height h reached by the jump. The height h is found by setting the kinetic energy from the jump to potential energy from the jump up.

KEjump = PEjump
(1/2)*m*(5.0m/s)^2 = mgh
Solve for h.

The work W = average spring force*d so
PE = W.
mg(H+h) = (1/2)*(6.2*10^4)*d*d

(I probably need to explain why the (1/2) term is there. The spring force increases from infinitesimal at the first infinitesimal deflection. At the max deflection the force is (6.2*10^4)*d. The average is (1/2)*(6.2*10^4)*d.)

Solve for d.
Hmmm, I just re-read the question. We don't know the height H? It's required.

Answer 3

tul b is sweet with area a, i will attempt to help with b. The peg differences the direction the ball follows partly a. From area a, while the twine makes touch with the peg, its speed is squarert. 2gL and its kinetic capability is KE = one million/2mv^2 = one million/2m2gL. It wraps around the peg and, while the ball is on the right of its around direction with reference to the peg, a number of that kinetic capability is switched over returned to capacity. call the kinetic area of the ball's capability above the peg KEp, and the capacity area while at that element PEp. the capacity capability PEp = mg*.4L (I had worry right here - the ball is .2L under the peg on the backside and swings to .2L above the peg. Delta top = .4L.) So KE = one million/2m2gL = KEp + PEp = KEp + mg*.4L So one million/2m2gL = KEp + mg*.4L KEp = one million/2m*Vpt^2 = one million/2m2gL - mg*.4L the place Vpt is speed on the right of the direction around the peg. Simplify Vpt^2 = 2gL - .8gL = one million.2gL Vpt = sqrt one million.2gL