WORKER ARE LOADING EQUIPMENT INTO A FREIGHT ELEVATOR AT THE TOP FLOOR OF A BUILDING HOWEVER THEY OVERLOAD THE ELEVATORAND THE WORN CABLE SNAPS THE MASS OF THE LOADED ELEVATOR AT THE TIME OF THE ACCIDENT IS 1600KG..AS THE ELEVATOR FALLS,THE GUIDE RAILS EXCERT A CONSTANT RETARDING FORCE OF 3700N ON THE ELEVATOR. AT WHAT SPEED DOES THE ELEVATOR HIT THE BOTOM OF THE SHAFT 72m BELOW?
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To find the speed at which the elevator hits the bottom of the shaft, we need to use the concept of energy.
1. First, let's calculate the potential energy of the elevator at the top floor using the formula:
Potential Energy = Mass * Acceleration due to gravity * Height
In this case, the mass of the loaded elevator is given as 1600 kg, the acceleration due to gravity is approximately 9.8 m/s², and the height of the shaft is given as 72 m.
Potential Energy = 1600 kg * 9.8 m/s² * 72 m
2. Next, let's calculate the work done against the retarding force of the guide rails. The work done is equal to the force multiplied by the distance traveled.
Work = Force * Distance
In this case, the force exerted by the guide rails is given as 3700 N, and the distance traveled is the same as the height of the shaft, which is 72 m.
Work = 3700 N * 72 m
3. Since the elevator is falling, we can assume that the initial kinetic energy is zero. Therefore, the final kinetic energy of the elevator will be equal to the initial potential energy minus the work done against the retarding force.
Final Kinetic Energy = Potential Energy - Work
4. The kinetic energy of an object is given by the formula:
Kinetic Energy = (1/2) * Mass * Velocity^2
Let's assume the velocity at the bottom of the shaft is V m/s.
Final Kinetic Energy = (1/2) * Mass * V^2
5. Now, we can equate the formulas for final kinetic energy and potential energy minus work.
(1/2) * Mass * V^2 = Potential Energy - Work
Substituting the values:
(1/2) * 1600 kg * V^2 = (1600 kg * 9.8 m/s² * 72 m) - (3700 N * 72 m)
6. Solve the equation to find the value of V, the velocity at the bottom of the shaft.
(1/2) * 1600 kg * V^2 = (1600 kg * 9.8 m/s² * 72 m) - (3700 N * 72 m)
Simplifying the right-hand side:
(1/2) * 1600 kg * V^2 = 1128960 J - 266400 J
(1/2) * 1600 kg * V^2 = 862560 J
V^2 = (2 * 862560 J) / (1600 kg)
V^2 = 1128.6 m^2/s^2
Taking the square root of both sides:
V = √1128.6 m^2/s^2
V ≈ 33.6 m/s
Therefore, the elevator hits the bottom of the shaft with a speed of approximately 33.6 m/s.