When the person has fallen between 8 m and the bottom. At the bottom, when the p

Question

When the person has fallen between 8 m and the bottom.

At the bottom, when the person has fallen 30 m.    

At the top, when the person has fallen 0 m.

When the person has fallen between 0 m and 8 m.

When the person has fallen 8 m.

(c) What is the jumper's speed at this instant, when the tension is greatest in the cords?
v = _____ m/s   

(e) Which of the following statements is a valid basis for answering part (d) correctly?

If the momentum weren't changing, the momentum would remain zero forever.

Since the momentum is zero, the momentum isn't changing.

A very short time ago the momentum was downward (and nonzero).

Since the net force must be zero when the momentum is zero, and since dpy/dt is equal to the net force, dpy/dt must be zero.

After a very short time the momentum will be upward (and nonzero).

(f) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness ks for each of the two cords.
ks = ____ N/m

(g) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.)
FT = ____ N

(h) What is the maximum acceleration |ay| = |dvy/dt| (in "g's") that the jumper experiences? (Note that |dpy/dt| = m|dvy/dt| if v is small compared to c.)
|ay| = _____ g's (acceleration in m/s2 divided by 9.8 m/s2)

(i) What is the direction of this maximum acceleration?

downward

upward    

no direction, since the acceleration is zero

(j) What approximations or simplifying assumptions did you have to make in your analysis which might not be adequately valid? (Don't check any approximations or simplifying assumptions which in fact have negligible effects on your numerical results.)

Assume that the gravitational force hardly changes from the top of the jump to the bottom.

Assume tension in cord proportional to stretch, even for the very large stretch occurring here.

Assume the speeds are very small compared to the speed of light.

Neglect air resistance, despite fairly high speeds.

Explanation / Answer

Measuring the heights from the final equilibrium position.

The potential energy at the top most point was mgh = 120*9.8*30 = 35280
All this energy is converted as the stings p.e at the bottom most point/
1/2 k x^2 = 1/2 k 22^2 = 35280
f)
ks = 145.78 N/m

g)
The force needed to extend 22 m is 145.78*22 = 3207 N
Assuming each string has the same spring constant, then each can support half the above value = 1603.50N

h)
The max upward acceleration a =1603.50/120 = 13.36 m/s^2.

i)

downward

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