How Is Potential Energy Converted to Kinetic Energy in Physical Systems?

[Sidhshah1234]

Homework Statement

A daredevil gazelle called Giselle takes her first bungee jump from a height of 160 m above the ground. Giselle’s mass is 26 kg, and when she is attached by the ankles to the bungee cord, her height means she will extend 2.0 m beyond the end of the bungee cord. Her bungee cord is 33 m long.

(a) How far does Giselle travel downward in freefall before the spring effect of the bungee stretching kicks in?[Ans: 33 m]

(b) Draw the free-body diagram of Giselle when the bungee spring force is acting.

(c) What was Giselle’s initial potential energy as she stepped off the platform at the top? [Ans: 41,000 J]

(d) Giselle’s bungee instructor has carefully chosen her bungee cord accounting for the platform height and Giselle’s weight, in order that she will just barely touch the ground. What is the spring constant, k, of the bungee cord?[Ans: 5.2 N/m]

(e) Recognizing that Giselle’s maximum velocity occurs when her acceleration is zero (she is at the point between speeding up and the bungee cord slowing her down again), how far has she fallen when she reaches this maximum velocity? [Ans: 82 m]

(f) What is her maximum velocity? [Ans: 34 ms-1]

I need help with d) - f)

Relevant Equations

F=kx
F=ma
PE=mgh
W=1/2kx^2

d)
W=1/2kx^2=PE
41000=1/2(k)(33)^2
k=82000/(33)^2
k=75.3N/m
(nowhere close to the answer)

e)
Potential Energy = Kinetic Energy
mgh = 0.5 * m * v^2

Given:
m = 26 kg
g = 9.8 m/s^2
h = 160 m

Solving for velocity (v):

0.5 * 26 kg * v^2 = 26 kg * 9.8 m/s^2 * 160 m
v^2 = (2 * 9.8 m/s^2 * 160 m)
v = sqrt(2 * 9.8 m/s^2 * 160 m) ≈ 56 m/s
v^2 = u^2 + 2as
0 = 0^2 + 2 * 9.8 m/s^2 * ss = 0.5 * (34 m/s)^2 / 9.8 m/s^2 ≈ 160m

 

[BvU]

Hello and !

(d) Giselle’s bungee instructor has carefully chosen her bungee cord accounting for the platform height and Giselle’s weight, in order that she will just barely touch the ground. What is the spring constant, k, of the bungee cord?[Ans: 5.2 N/m]


I need help with d) - f)
Relevant Equations: F=kx
F=ma
PE=mgh
W=1/2kx^2

41000=1/2(k)(33)^2

Did you make a sketch ? Please post it ...

What is this $x=33$ on the sketch ?

$\$

 

[BvU]

For the record:

• what value of $g$ are you supposed to use ? Ah, I see a 9.8 m/s² in your part e) answer (*)
• I have trouble with the book answer for part a). If
  

  she stepped off the platform at the top

  and the 33 m answer is correct she is still upright !?
• The problem statement gives you the girl's length. How do you use that ? (my answer: you seem to ignore it. But: What if she was 20 m high instead of 2 ? )

(*):

e)
Potential Energy = Kinetic Energy
mgh = 0.5 * m * v^2

Given:
m = 26 kg
g = 9.8 m/s^2
h = 160 m

Solving for velocity (v):

0.5 * 26 kg * v^2 = 26 kg * 9.8 m/s^2 * 160 m
v^2 = (2 * 9.8 m/s^2 * 160 m)
v = sqrt(2 * 9.8 m/s^2 * 160 m) ≈ 56 m/s
v^2 = u^2 + 2as
0 = 0^2 + 2 * 9.8 m/s^2 * ss = 0.5 * (34 m/s)^2 / 9.8 m/s^2 ≈ 160m

This looks a lot like a disaster where someone forgot the bungee cord


$\$

 

[Herman Trivilino]

W=1/2kx^2=PE
41000=1/2(k)(33)^2

You are saying the bungee cord stretches 33 m?

 

[jbriggs444]

Homework Statement: A daredevil gazelle called Giselle takes her first bungee jump from a height of 160 m above the ground. Giselle’s mass is 26 kg, and when she is attached by the ankles to the bungee cord, her height means she will extend 2.0 m beyond the end of the bungee cord. Her bungee cord is 33 m long.

(a) How far does Giselle travel downward in freefall before the spring effect of the bungee stretching kicks in?[Ans: 33 m]

What about Giselle's height? The problem setter wants us to consider it.

As I see it, Giselle is about 2 meters in height. Her center of gravity begins about 1 meter above her ankles (161 meters above the ground). During the head-first fall, her center of gravity is about 1 meter below her ankles.

 

[erobz]

The question setter apparently also thinks it’s a good idea that the bungee operator chooses a bungee such that her head just barely touches the ground in part d…

 

[haruspex]

and the 33 m answer is correct she is still upright !?

She could well be still upright at the instant the cord becomes taut. Her orientation would change abruptly after that.