Work and acceleration of rescue team

ViewtifulBeau · Sep 24, 2005

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in in three stages, each requiring a vertical distance of 11.1 m: (1) the initially stationary spelunker is accelerated to a speed of 4.06 m/s; (2) he is then lifted at the constant speed of 4.06 m/s; (3) finally he is decelerated to zero speed. How much work is done on the 69.4 kg rescue by the force lifting him during each stage?

I have to find how much work is done by the force lifting him at each stage.

First I found the average velocity 4.06/2 = 2.03 (i think because of asuming constant acceleration). Then i divide the distance by the average velocity to get 11.1 / 2.03 = 5.4679 s . So this is how long it takes to make it to the end of the 1st stage. Then i found the acceleration which is .7425 m/s/s. but i plugged it into the work formula and i got 572 J... but it is wrong.

zwtipp05 · Sep 24, 2005

Each step is done in 11.1 m, so that makes it easier.
W=F times distance.
In each step the distance will be 11.1 m.

Step 1: 11.1 = .5at^2 a=(v_f-v_i)/t
Solve for a. Then consider what force is needed to get that acceleration (keep in mind they are working AGAINST gravity as well)

Step 2: a=0 so force is easy to figure (again, keep in mind gravity)

Step 3: acceleration is the same as step 1 except opposite in magnitude, so they are letting gravity help them here)

Hope this helps

lightgrav · Sep 25, 2005

The key benefit of Work and Energy approaches
is that you get to ignore the details of the motion
...like acceleration.
The Work done in part 1 increases PE and KE,
so W = m g Delta_h + 1/2 m v^2 .
Work in Part 2 only has the Delta_PE
Work in part 3 is Delta_PE - 1/2 m v^2 ,
since the KE_final - KE_initial = - 1/2 m v^2 .

zwtipp05 · Sep 25, 2005

lightgrav said:

The key benefit of Work and Energy approaches
is that you get to ignore the details of the motion
...like acceleration.
The Work done in part 1 increases PE and KE,
so W = m g Delta_h + 1/2 m v^2 .
Work in Part 2 only has the Delta_PE
Work in part 3 is Delta_PE - 1/2 m v^2 ,
since the KE_final - KE_initial = - 1/2 m v^2 .

touche, didn't think of that

ViewtifulBeau · Sep 25, 2005

ok i tried the equations and i got 7700, 7557, and 7414 J respectively. The 7557 worked but the other two didn't. So i think that my v value is wrong. I used 2.03 m/s because that would be the average. But should i use 4.06m/s?

Päällikkö · Sep 25, 2005

Yes, you should use the final velocity: That's how much kinetic energy the system gets.