Question

You need to design a safety device for a temporary lift . The device can be modeled as a spring below the elevator such that if the cable holding the elevator car breaks, the spring will slow down and stop the car. Your design needs to meet the following conditions: - The elevator car has a mass m =667 kg - The elevator car starts at a height yᵢ =3.45 m above the top of the spring. - The elevator car is initially moving upwards with a velocity vₜ =4.53 m/s - The elevator car needs to be stopped a distance y =1.76 m below the original position of the end of the spring. - You can neglect friction and air drag in your design. What is the spring constant needed for the spring that will meet the design?

Answer 1

To calculate the spring constant needed for the safety device, we can use the principle of conservation of energy. The formula for the spring constant is k = (2mg(y_initial - y_final)) / (y_final)^2.

Step-by-step explanation:

To calculate the spring constant needed for the safety device, we can use the principle of conservation of energy. Initially, the elevator car has gravitational potential energy, which is converted into the potential energy stored in the compressed spring when the cable breaks.

The gravitational potential energy can be calculated using the equation PE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the initial height of the car above the top of the spring.

The potential energy stored in the compressed spring is given by the equation PE = 0.5k(yfinal)2, where k is the spring constant and yfinal is the distance the car needs to be stopped below the original position of the spring.

Setting the two equations equal to each other and solving for k, we get k = (2mg(yinitial - yfinal)) / (yfinal)2.