Required information You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of $2400 \mathrm{~kg}$ (including passengers). For passenger comfort, you choose the maximum ascent speed to be 18.0 ms, the maximum descent speed to be $10.0 \mathrm{~m} / \mathrm{s}$, and the maximum acceleration magnitude to be $5.40 \mathrm{~m} / \mathrm{s}^{2}$. Ignore friction. What is the maximum upward force that the supporting cables exert on the elevator car? $\mathrm{kN}$

Step 1 ：Given that the mass of the elevator is 2400 kg, the acceleration due to gravity is 9.8 m/s², and the maximum acceleration of the elevator is 5.4 m/s².

Step 2 ：First, calculate the weight of the elevator using the formula \(Weight = mass \times gravity\). Substituting the given values, we get \(Weight = 2400 \times 9.8 = 23520.0 N\).

Step 3 ：Next, calculate the force required to accelerate the elevator upwards using the formula \(Force = mass \times acceleration\). Substituting the given values, we get \(Force = 2400 \times 5.4 = 12960.0 N\).

Step 4 ：The total force exerted by the supporting cables is the sum of the weight of the elevator and the force required to accelerate it upwards. So, \(Total Force = Weight + Force = 23520.0 + 12960.0 = 36480.0 N\).

Step 5 ：Since 1 kN = 1000 N, we can convert the total force to kN by dividing by 1000. So, \(Total Force = \frac{36480.0}{1000} = 36.48 kN\).

Step 6 ：Final Answer: The maximum upward force that the supporting cables exert on the elevator car is \(\boxed{36.48}\) kN.