Spring and falling body problem

[professor]

1. Neglect air resistance. A suspention bridge spand the colorado river at a height of 321 meters. Consider a 75kg bungee jumper, and a 200m bungee cord which obeys hooke's law (once it reaches its unstreched length).

a) calculate the spring constant k, so that the jumper halts just above the surface
b) What is the magnitude and direction of the jumper’s acceleration at the bottom of the jump? Express in G's.
c) Find the height above the river where the jumper reaches maximum speed.
d) Find the maximum speed. Terminal velocity for a falling human is about 120 mph.
Did the jumper reach terminal velocity? If so, we should take into account air resistance but that’s another problem).



2. F=-kx, x=.5at² (x initial is 0, v initial is 0), x=x_i+v_it+.5at² , V=at



3. If I write out less of an answer, it is because I think I have it correct. If I do not, I will certainly write everything out in more detail.

Answers:
a) I set the force of gravity equal to the force of the spring for position 321 meters, but since the spring doesn't do anything until 200m are passed, I actually used 121 meters as the x point for the spring. Just a little algebra got me a spring constant of about 6 kg/s², which I believe is correct, until I use it later.

Here is what I did:
mg=-kx
75kg*9.8m/s²=-k*121m
this give me k= -6.07kg/s²

Is there anything wrong with this?

b) At the bottom of the jump, it seems to me that the forces are canceled (the force of the spring versus the force of gravity). Which would mean Acceleration is 0. This seems intuitively correct, but my intuition is not allways correct.

c) I did all sorts of things with this problem, which I wish I could scan. Basically, I haven't really got anywhere, but here are some key points of my attempt:
The jumper has a=-9.8m/s² until the point 200 meters. after this point, I think the accel is: -9.8 + (6/76kg)*(x-200). I calculated the acceleration added by the spring by setting -kx = ma, and solving for a, where I used 6kg//s² for k. I tried to use the position formula: x=x_i+v_it+.5at², where x_i was equal to 200m, v_i was equal to 62.6m/s (which I calculated was the velocity of the jumper at position equals 200m), and a=-9.8 + (6/76kg)*(x-200). Then i basically got stuck. I can't figure out the time or height of the jumper at his max speed. Presumably this would be when a=0. Perhaps a=o when I solve for my equation for a, but this gives me x=77.5 meters from the start, which makes no sense because no force is acting against the acceleration of gravity until atleast 200 meters.

This probably is a result of my spring coefficient being wrong.

d) Once I find out where the max speed occurs, I will be able to easily calculate this I think.

 

[alphysicist]

Hi professor,

1. Neglect air resistance. A suspention bridge spand the colorado river at a height of 321 meters. Consider a 75kg bungee jumper, and a 200m bungee cord which obeys hooke's law (once it reaches its unstreched length).

a) calculate the spring constant k, so that the jumper halts just above the surface
b) What is the magnitude and direction of the jumper’s acceleration at the bottom of the jump? Express in G's.
c) Find the height above the river where the jumper reaches maximum speed.
d) Find the maximum speed. Terminal velocity for a falling human is about 120 mph.
Did the jumper reach terminal velocity? If so, we should take into account air resistance but that’s another problem).



2. F=-kx, x=.5at² (x initial is 0, v initial is 0), x=x_i+v_it+.5at² , V=at



3. If I write out less of an answer, it is because I think I have it correct. If I do not, I will certainly write everything out in more detail.

Answers:
a) I set the force of gravity equal to the force of the spring for position 321 meters, but since the spring doesn't do anything until 200m are passed, I actually used 121 meters as the x point for the spring. Just a little algebra got me a spring constant of about 6 kg/s², which I believe is correct, until I use it later.

Here is what I did:
mg=-kx
75kg*9.8m/s²=-k*121m
this give me k= -6.07kg/s²

Is there anything wrong with this?

I don't believe this is correct. The forces are not canceled at the point right above the water. Since the jumper's speed is zero there, if all of the forces canceled then he would stay there indefinitely; but in fact we know that the cord will pull him back upwards (so at that point the spring force must be greater than the gravitational force).

I would suggest you think in terms of energy for this part. What does that give?

 

[baba89]

Your approach to the problem is on the right track. It is important to note that in this problem, the bungee cord is not stretched until the jumper reaches the 200m mark. This means that until that point, the force of the spring is not acting on the jumper and the only force acting is gravity.

a) To calculate the spring constant, we can use the equation F = kx, where F is the force of the spring, k is the spring constant, and x is the displacement from the unstretched length of the spring. Since the jumper stops just above the surface, we can set the force of the spring equal to the force of gravity at a height of 321m. This gives us the following equation:

mg = kx

Substituting in the values given, we get:

(75kg)(9.8m/s^2) = k(121m)

Solving for k, we get a spring constant of approximately 6.07 kg/s^2. This is correct, as long as we use it consistently throughout the problem.

b) At the bottom of the jump, the forces are indeed cancelled out and the acceleration is 0. This is because the bungee cord is no longer stretched and there is no longer any opposing force from the spring. The jumper will experience a brief moment of weightlessness before being pulled back up by the bungee cord.

c) To find the height where the jumper reaches maximum speed, we can use the equation for displacement using constant acceleration:

x = xi + vit + 0.5at^2

At the point where the jumper reaches maximum speed, the acceleration will be 0. So we can set the equation equal to 0 and solve for t. This will give us the time it takes for the jumper to reach maximum speed. We can then plug this value into the equation to find the height.

d) Once we have the height and time at maximum speed, we can use the equation for velocity using constant acceleration to find the maximum speed. If the maximum speed is greater than the terminal velocity (120 mph), then the jumper did not reach terminal velocity and air resistance would need to be taken into account. If the maximum speed is less than 120 mph, then the jumper did reach terminal velocity and air resistance can be neglected in this problem.

Overall, your approach and calculations are correct. Just make sure to use the correct values for the