An elevator in a tall building is allowed to reach a maximum speed of 3.0 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.4 m if the elevator has a mass of 1500 kg including occupants? Need step by step help please!
find the accelearation:
Vf^2=Vi^2+2ad
a= -1/2 Vi^2/d= in my head....-4.5/2.4
tension= mass(g+4.5/2.4)
To calculate the tension in the cable required to stop the elevator, we need to use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force causing the elevator to come to a stop is the tension in the cable.
Step 1: Find the acceleration
To find the acceleration, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s as the elevator comes to a stop)
u = initial velocity (3.0 m/s as given in the question)
s = distance traveled (2.4 m as given in the question)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
= (0^2 - 3.0^2) / (2 * 2.4)
Step 2: Calculate the acceleration
Substituting the values:
a = (-9.0) / (4.8)
Step 3: Calculate the net force
Now that we have the acceleration, we can calculate the net force:
F = ma
where:
m = mass of the elevator (1500 kg as given in the question)
a = acceleration calculated in Step 2
Substituting the values:
F = 1500 kg * (-9.0 / 4.8) m/s^2
Step 4: Calculate the tension
The tension in the cable is equal to the net force acting on the elevator:
T = F
Substituting the values:
T = 1500 kg * (-9.0 / 4.8) m/s^2
Calculating the result will give you the tension required to stop the elevator over a distance of 2.4 m.