thought this was a springs question....not sure now

[Addallat]

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A 100 kg bungee jumper attached to a bungee cord jumps off a bridge. The bungee cord stretches and the man reaches the lowest spot in his descent before beginning to rise. The force of the stretched bungee cord can be approximated using hooke's law, where the value of the spring constant is replaced by an elasticity constant, in this case 100 kg/s^2. If the cord is stretched by 30 m at the lowest spot of the man's decent, then what is his acceleration at the lowest spot?



A. 0 m/s^2

B. 10 m/s^2

C. 20 m^/s^2

D. 30 m/s^2


The way I went about solving this problem was completely wrong, and the explanation in the book just had me even more confused. Could someone out there write out how they solve this problem, thanks. Oh and the answer is c.

 

[reising1]

A 100 kg bungee jumper attached to a bungee cord jumps off a bridge. The bungee cord stretches and the man reaches the lowest spot in his descent before beginning to rise. The force of the stretched bungee cord can be approximated using hooke's law, where the value of the spring constant is replaced by an elasticity constant, in this case 100 kg/s^2. If the cord is stretched by 30 m at the lowest spot of the man's decent, then what is his acceleration at the lowest spot?



A. 0 m/s^2

B. 10 m/s^2

C. 20 m^/s^2

D. 30 m/s^2


The way I went about solving this problem was completely wrong, and the explanation in the book just had me even more confused. Could someone out there write out how they solve this problem, thanks. Oh and the answer is c.

You'll want to draw out a picture. Draw a square - representing the man on the bungee in mid air - and think about what forces are acting on it. There is the weight of the man (W = mg) and then there is the force on the bungee (F = -kx). The sum of the forces in the Y direction equals mass * acceleration (Newton's 2nd Law).

W + F = ma
mg - kx = ma
(100 kg)(10 m/s^2) - (100 kg/s^2)(30 m) = (100 kg)(a)
a = -20 m/s^2

The negative here indicates the man will begin accelerating upwards. Note that this makes sense, because the man cannot accelerate downwards faster than 10 m/s^2 (gravity), unless someone pushed him down or something like that. But since he jumped on his own, 10 m/s^2 downwards is the fastest acceleration he can achieve.

Let me know if that helps!

 

[discowisco]

EDIT: ^ above answer is more clear haha

 

[Addallat]

ahhhhhhh thank you for spelling that out! i knew i was making a stupid mistake somewhere; i kept taking -k x^2 for hooke's law instead of -kx

 

[reising1]

ahhhhhhh thank you for spelling that out! i knew i was making a stupid mistake somewhere; i kept taking -k x^2 for hooke's law instead of -kx

Ah, yeah, don't get confused about elastic mechanics.

Force = -kx
Potential Energy = (1/2)kx^2