# Work and acceleration of rescue team

ViewtifulBeau
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
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

Homework Helper
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 .

Last edited:
zwtipp05
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
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?

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