Question

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 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?

Answer 1

(a) 9305 J

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

`v^2-u^2=2ad`

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,

`a=(v^2-u^2)/(2d)=((2.40 m/s)^2-0)/(2(11.0 m))=0.26 m/s^2`

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

`F-mg = ma`

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,

`F=m(a+g)=(84.0 kg)(0.26 m/s^2+9.81 m/s^2)=845.9 N`

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

`W=Fd=(845.9 N)(11.0 m)=9305 J`

(b) 9064 J

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

`F-mg=0`

From which we find

`F=mg=(84.0 kg)(9.81 m/s^2)=824.0 N`

And so the work done is

`W=Fd=(824.0 N)(11.0 m)=9064 J`

(c) 8824 J

The acceleration of the spelunker here is given by

`v^2-u^2=2ad`

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,

`a=(v^2-u^2)/(2d)=(0-(2.40 m/s)^2)/(2(11.0 m))=-0.26 m/s^2`

Newton's second law applied to the spelunker is:

`F-mg = ma`

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,

`F=m(a+g)=(84.0 kg)(-0.26 m/s^2+9.81 m/s^2)=802.2 N`

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

`W=Fd=(802.2 N)(11.0 m)=8824 J`