What is the total length of the bungee cord in this bungee jumping problem?

[robax25]

Homework Statement

A bungee jumper weight 60 kg, spring constant 98N/m, the cord is extended by 18m.Natural length of the cord is 9m.total length is 27m.

max force,
F=kx
mg=kx
x=6m.
it goes to 6m extension but to extend more from where it gets extra force?

Homework Equations

F=kx

The Attempt at a Solution

x=6m[/B]

 

[jbriggs444]

Homework Statement

A bungee jumper weight 60 kg, spring constant 98N/m, the cord is extended by 18m.Natural length of the cord is 9m.total length is 27m.

This does not sound like a complete statement of the problem. In particular, it does not ask a question.

 

[robax25]

its total extension is 18m and its equilibrium length is 6m. my question is that, why does extend it more than 6m?

 

[jbriggs444]

its total extension is 18m and its equilibrium length is 6m. my question is that, why does extend it more than 6m?

At an extension of 6m [beyond the unstressed 9m], what is the net force on the bungee jumper?

 

[robax25]

Net force = kx-mg=98N/m*6m-60kg*9,8m/s²=0.6N

 

[jbriggs444]

Net force = kx-mg=98N/m*6m-60kg*9,81m/s²=0.6N

Why are you using two different values for g in the same formula?

 

[robax25]

the net force would be zero.

 

[jbriggs444]

g=9.81m/s²

OK, I jumped the gun a bit on that reply. The spring constant is clearly chosen to have a numerical value equal to 10g so that the equilibrium position comes out at 6m exactly. Your answer of 0.6N correctly reflects the fact that the two numerical values are not an exact match.

0.6N is small. What does that mean for the acceleration of the jumper at 6m extension?

 

[robax25]

I mean that they (spring force and gravitational force) are now equilibrium at 6m.

 

[jbriggs444]

I mean that they (spring force and gravitational force) are now equilibrium at 6m.

Right. So the net force is zero. What does that say about the acceleration at that point?

 

[robax25]

acceleration 9.8m/s²

 

[jbriggs444]

acceleration 9.8m/s²

Come again? The net force is zero and already includes gravity. F=ma.

What is the acceleration?

 

[robax25]

a=0

 

[jbriggs444]

a=0

Right.

Now consider what happens as the bungee jumper is falling toward this equilibrium point at 6m. Is he accelerating? In which direction?

 

[robax25]

he is accelerating downward

 

[jbriggs444]

he is accelerating downward

So he will have a non-zero velocity when he reaches the equilibrium point, yes?

 

[robax25]

yes

 

[robax25]

acceleration is zero but velocity is not

 

[jbriggs444]

yes

Which means that he is going to keep on falling past it. Which is, I think, what you were trying to justify.

 

[robax25]

yes I would like to ask you a question from force oscillation. you are really good in fundamental thing

 

[robax25]

also a good teacher.

 

[robax25]

A driver drives along a road .The road has periodic small bump of 5 cm height and a distance of 5 cm. Car's shock observer works fine,damping the deflection to half each oscillation.spring length is 20cm increased if car weight(1500kg) does not act.Determine the oscillation amplitude if the car is driving at 20km/h.

consider the problem by assuming that the distance of the four wheels equals multiple of the periodicity.

 

[jbriggs444]

A driver drives along a road .The road has periodic small bump of 5 cm height and a distance of 5 cm. Car's shock observer works fine,damping the deflection to half each oscillation.spring length is 20cm increased if car weight(1500kg) does not act.Determine the oscillation amplitude if the car is driving at 20km/h.

consider the problem by assuming that the distance of the four wheels equals multiple of the periodicity.

I do not understand the question.

First of all, it is unrealistic. With 5cm height bumps at 5 cm separation, the car tires will prevent the depth of the grooves from being at all relevant. This means that 90% of the problem statement is rubbish.

Second, we are told that the oscillations are reduced to half value if the car is massless. But that is idiotic. A massless car will not have its motion damped at all. It will ideally rise and fall exactly with the bumps.

However, a hypothetical massless car doing this at 20 km/h over 5 cm high, 5 cm spacing bumps would be experiencing negative g's (without even putting pencil to paper, this seems obvious). So realism demands that the car be in free fall much of the time with the wheels off the ground. That makes the problem much more complicated.

So I am at a loss. I do not know what knowledge this question is trying to probe.

 

[robax25]

it is not massless. it compresses 20cm when force acts(1500kg). The wording is wrong. For example, you applied weight a force on a spring, it goes down 20cm and then, you relieve the force, then, it goes up. Bump size 5cm*5cm is not impossible. length and Height. it is like a rectangle.

 

[jbriggs444]

Bump size 5cm*5cm is not impossible

It is not impossible -- but tires cannot fit. Which makes the problem depend on tire diameter.

And without 1500 kg of body weight, there is nothing to resist the force of the spring. The attached object will simply trace out the profile of the track.

 

[robax25]

However, here is only need the amplitude of driving oscillation. If you drive a car, you spring compresses 20cm due to car weight and you face bump(20cm*20cm) periodic, your car is running 20km/h

 

[jbriggs444]

However, here is only need the amplitude of driving oscillation.

But we are told, counter-factually, that the resulting oscillation is different. Either I am missing something or the problem statement is. [And we still have the problem that the driving oscillation is not specified in the problem statement]