Question

g A person exploring a deep cave system becomes injured and needs to be rescued. The fastest way to get them is to pull them straight up out of the cave through a small opening just overhead, using a motor-driven cable. The lift is performed in three stages, each of them 10 m in height (total of 30 meters to extract the person). In the first stage, the person is accelerated to a speed of 5 m/s. They are then lifted at constant speed of 5 m/s, then in the last stage they are slowly decelerated to zero speed. If the person weighs 80 kg, how much work is done in each stage

Answer 1

Answer:

1. W = 8848 J

2. W = 7848 J

3. W = 6848 J

Explanation:

The work (W) can be found using the following equation:

$W = E_{k} + E_{p}$

Where: E(k) is the kinetic energy and E(p) is the potential energy

Now let's find the work for every stage.

Stage 1:

$W = E_{k} + E_{p} = \frac{1}{2}mv^{2} + mgh$

Where: m is the mass, g is the gravity, h is the height, v is the speed  

$W = \frac{1}{2}mv^{2} + mgh = \frac{1}{2}80 kg*(5 m/s)^{2} + 80 kg*9.81 m/s^{2}*10 m = 8848 J$

Stage 2:

$W = E_{k} + E_{p} = 0 + E_{p}$

The kinetic energy is equal to zero because the acceleration is constant.

$W = E_{p} = mgh = 80 kg*9.81 m/s^{2}*10 m = 7848 J$

Stage 3:

$W = E_{k} + E_{p} = \frac{1}{2}mv^{2} + mgh = -\frac{1}{2}80 kg*(5 m/s)^{2} + 80 kg*9.81 m/s^{2}*10 m = 6848 J$      

I hope it helps you!