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 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 4.90 m/s; (b) he is then lifted at the constant speed of 4.90 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 68.0 kg rescue by the force lifting him during each stage?

Answer 1

Let's agree

$m=68kg$ mass of spelunker

$s=10m$ - distance for each stage

$a-$ acceleration

$t-$ time of movement

$g=9.8 m/s^2$

For the srage (a):

$v_0=0$

$v=4.9m/s$ - speed in the end of stage (a)

The equation of movement for stage (a):

$$v=v_0+(a-g)t$$ $$s=v_0t+(a-g)t^2/2$$ $$\implies$$ $$a=g+v^2/2s$$

Using formula for the work made by force and Newton's second law :

$$A=Fs$$ $$F=ma$$ $$A_a=m(g+v^2/2s)s$$

For the stage (b) with the constant velocity acconding the Newton's second and third laws

$$F=ma=mg$$

Using formula for the work made by force

$$A_b=Fs=mgs$$

The equation of movement for stage (c):

$$v=v_0+(a-g)t$$ $$s=v_0t+(a-g)t^2/2$$

in this case $v=0$ - movement until stop, $v_0=4.9m/s$ - start moving speed

$$\implies$$ $$a=g-v_0^2/2s$$

Using formula for the work made by force and Newton's second law :

$$A=Fs$$ $$F=ma$$ $$A_c=m(g-v_0^2/2s)s$$

The total work:

$$A=A_a+A_b+A_c$$

Using numbers:

$$A=19992 J$$

Answer: $A=19992 J$