Rock climbing energy conservation

[dfx]

Hi,
A rather simple question that I just can't seem to solve:

The greatest instantaneous acceleration a person can survive is 25g, where g is the acceleration of free fall. A climber's rope should be selected such taht, if the climber falls when the rope is attached to a fixed point on a vertical rock, the fall will be survived.

A climber of mass m is attached to a rope which is attached firmly to a rock face at B as shown. When at a point A, a distance L above B, the climber falls.

http://img144.imageshack.us/img144/2483/rockclimbjh9.jpg

(the red bit is the rope)

(a) Assuming that the rope obeys Hooke's law upto the breaking, use the principle of conservation of energy and the condition for greatest instantaneous acceleration to show that the part AB of the rope (of unstretched length L) must be able to stretch by more than L/6 whithout breaking for the climber to survive

My working:

After the climber falls under freefall for a maximum distance of 2L (from A to B, and then the length L of the rope which makes it a total of 2L), then he has maximum kinetic energy which is K.E = (1/2)mv^2 but v^2 = u^2 + 2as = 4gL

Then K.E. = elastic potential energy + gpe ... (i)

EPE = (1/2)kx^2 = (1/2)Fx

and GPE = mgx , where x is the extension during the rope stretching.

So from (i):

(1/2)mv^2 = mgx + (1/2)Fx and since F = ma, with the maximum instantaneous acceleration of 25g, then F = 25mg

2mgL = mgx + (25/2)mgx

2L = (27/2)x

Therefore x = (4/27) L ? but the required answer is x = L/6 ?

 

[Delzac]

here's my suggestion, don't try and kill yourself by having expression like:

EPE = KE + GPE

KE after all comes from GPE so y dun we express it as:

EPE = GPE(Big)

Makes life much easier isn't?

 

[dfx]

But I don't get how you can just exclude the KE that is gained when falling the distance of 2L, because the GPE I've included is purely for the extension bit (x), and does not overlap with the GPE during the 2L fall.

Also if you do exclude it as you suggest, then I can't seem to get an equation with "L" in it, which is fundamental to solving the q.

Thanks for your help though.

 

[Andrew Mason]

So from (i):

(1/2)mv^2 = mgx + (1/2)Fx and since F = ma, with the maximum instantaneous acceleration of 25g, then F = 25mg

2mgL = mgx + (25/2)mgx

2L = (27/2)x

Therefore x = (4/27) L ? but the required answer is x = L/6 ?

The change in gravitational potential energy is \Delta(PE) = mg(2L+\Delta x) where \Delta x is the amount of the rope stretch. This is all converted into elastic potential energy of the rope at maximum stretch (KE = 0). If the maximum deceleration is 25g, the force is 26mg (weight of climber + 25mg for the deceleration). So kx = F has a maximum of 26mg:

mg(2L+\Delta x) = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}26mg(\Delta x)

AM

 

[dfx]

Ah I ignored the weight! Thank you.

Also seem to have got the energy slightly messed up but I understand now.

 

[Clare Taylor]

thanks i was stuck on this queation and missed the weight of the person as well!