Acave 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 three stages, each requiring a vertical distance of 11.0 m: (a) the initially stationary spelunker is accelerated to a speed of 2.40 m/s; (b) he is then lifted at the constant speed of 2.40 m/s; (c) finally he is decelerated to zero speed. how much work is done on the 84.0 kg rescue by the force lifting him during each stage?

(a) 9305 J

Let's start by finding the acceleration of the spelunker, through the following equation:

where

v = 2.40 m/s is the final velocity

u = 0 is the initial velocity

a is the acceleration

d = 11.0 m is the distance covered

Solving for a,

Now we can find the force lifting the spelunker. The equation for Newton's second law applied to the spelunker is:

where

F is the lifting force

m = 84.0 kg is the mass of the spelunker

g = 9.81 m/s^2 is the acceleration due to gravity

a = 0.26 m/s^2 is the acceleration

Solving for F,

And now we can finally find the work done on the spelunker by the lifting force F:

(b) 9064 J

In this case, the speed is constant, so the acceleration is zero. So Newton's second Law becomes

From which we find

And so the work done is

(c) 8824 J

The acceleration of the spelunker here is given by

where

v = 0 is the final velocity

u = 2.40 m/s is the initial velocity

a is the acceleration

d = 11.0 m is the distance covered

Solving for a,

Newton's second law applied to the spelunker is:

where

F is the lifting force

m = 84.0 kg is the mass of the spelunker

g = 9.81 m/s^2 is the acceleration due to gravity

a = -0.26 m/s^2 is the acceleration

Solving for F,

And now we can finally find the work done on the spelunker by the lifting force F: