Sure, let's solve this problem step-by-step.
1. Known Values and Assumptions:
- Mass of the elevator (m): 800 kg.
- Acceleration due to gravity (g): 9.81 m/s².
- Each floor height: 20 meters.
- Maximum velocity (v_max): 5 m/s.
2. Calculate the Total Distance from the 10th to Halfway between the 6th and 7th Floor:
- From the 10th floor to the 5th floor:
[tex]\[
5 \text{ floors} \times 20 \text{ meters/floor} = 100 \text{ meters}
\][/tex]
- The cab starts braking halfway between the 6th and 7th floors, so we reduce the total distance by:
[tex]\[
0.5 \text{ floors} \times 20 \text{ meters/floor} = 10 \text{ meters}
\][/tex]
- Thus, the total distance for braking starting is:
[tex]\[
100 - 10 = 90 \text{ meters}
\][/tex]
This "total distance" is 90 meters.
3. Calculate Braking Distance:
- The point where braking starts is halfway between the 6th and 7th floors, which is:
[tex]\[
(6 \text{ floors} \times 20 \text{ meters/floor}) - 10 \text{ meters} = 120 - 10 = 110 \text{ meters}
\][/tex]
- The total distance calculated to be 90 meters, and the braking starts at 110 meters down:
[tex]\[
90 \text{ meters} - 110 \text{ meters} = -20 \text{ meters}
\][/tex]
This negative distance essentially means the cab is moving upwards in our context. So, braking distance is -20 meters.
4. Determine the Kinetic Energy:
[tex]\[
\text{Kinetic energy} = 0.5 \times \text{mass} \times (\text{velocity})^2
\][/tex]
[tex]\[
= 0.5 \times 800 \times (5)^2 = 0.5 \times 800 \times 25 = 10000 \text{ joules}
\][/tex]
5. Calculate the Braking Force:
[tex]\[
\text{Work done by brakes} = \text{Force} \times \text{Braking distance}
\][/tex]
Rearranging, we solve for force:
[tex]\[
\text{Force} = \frac{\text{Kinetic energy}}{\text{Braking distance}} = \frac{10000}{-20} = -500 \text{ newtons}
\][/tex]
The negative sign indicates that braking force is in the direction opposite to the motion.
6. Adjust for Effective Force due to Cable Tension:
- Given the tension ratio (Tension when brakes are off is twice the tension when brakes are on), a ratio of 2.
[tex]\[
\text{Effective force} = \frac{\text{Force braking}}{\text{Tension ratio}} \times \text{mass} \times g
\][/tex]
[tex]\[
= \frac{-500}{2} \times 800 \times 9.81
\][/tex]
[tex]\[
= -250 \times 800 \times 9.81
\][/tex]
[tex]\[
= -1962000 \text{ newtons}
\][/tex]
This considers the combined effect of tension and gravitational counterforce.
So the elevator's brakes exert approximately [tex]\(-1,962,000 \, \text{newtons}\)[/tex] of force.
