A 2000 kg elevator is at rest when its cable breaks the...

A 2000-kg elevator is at rest when its cable breaks. The elevator falls 26 m before it encounters a giant spring at the bottom of the elevator shaft. As the elevator falls, its safety clamp applies a constant frictional force of 17,000 N. 1. List all the forces acting on the elevator after the cable breaks but before it hits the spring. Determine the work done by each force on the elevator. Make sure these work terms have the appropriate signs (positive or negative). 2. Calculate the net work done on the elevator. 3. Use the work-kinetic-energy theorem to determine the speed of the elevator right before it encounters the spring. 4. After the elevator encounters the spring, the spring compresses by 0.6 m before the elevator comes to rest. The safety clamp continues to apply a constant 17,000 N force. List all the forces acting on the elevator while it falls this final 0.6 m. 5. For each force listed in question 4, determine an expression for the work it does on the elevator. Make sure each work term has the appropriate sign (positive or negative). 6. Use the work-kinetic-energy theorem to determine the value of the spring constant k.

Transcript

00:01 In this example, i'm going to be looking at the relationship between workforce, distance, and energy.

00:08 Okay, so what we have is an elevator in an elevator shaft.

00:13 Okay, this elevator has a mass.

00:16 Elevator car has a mass m of 2 ,000 kilograms.

00:26 It's held up by a cable, but the cable suddenly snaps.

00:31 Okay, and the elevator car starts plummeting to a distance of h equals 26 meters before it will come in contact with a spring at the bottom of the shaft.

00:49 Okay, so this distance here is my h.

00:54 Okay, as the elevator is falling down the shaft, there are safety clamps that are providing a constant frictional force.

01:02 F sub f of 17 ,000 newtons.

01:08 Okay, and we're going to be looking at the forces on this elevator car as it falls.

01:16 The forces when it comes in contact with the spring and the work done by these various forces.

01:22 And then we're going to find the elevator's velocity just before it hits the spring.

01:27 And then we're going to find the spring constant k for this spring.

01:30 Okay, so the first thing is we want to identify the forces on this elevator car as it moves down the shaft.

01:37 Okay, and first of all, i'm going to define a coordinate system.

01:41 I'm going to call the downward direction positive.

01:44 Upward force will be negative.

01:47 Okay.

01:48 So the forces that i have acting on this elevator car are its weight.

01:54 So we have mass times acceleration due to gravity.

01:58 Okay, and that's in the downward direction.

02:00 That's positive.

02:03 And then we have the frictional force which opposes the direction of motion.

02:08 So that's going to be in the upward direction.

02:10 So that's negative.

02:12 Okay.

02:13 And that equals the sum of all forces acting on the car.

02:18 That's also going to equal mass times acceleration.

02:22 Okay, now we want to find the work done by each of these forces.

02:25 So that's simply the force times the distance over which it acts.

02:29 Okay, so that's going to be our h value 26 meters.

02:34 Okay, so my first term is going to be my weight times my distance.

02:44 Okay, so that's going to be m, g, h, and that equals 5 .10 times 10 to the 5 joules.

02:59 Minus the work done by friction.

03:02 So that's my frictional force times h.

03:07 Okay.

03:08 And that term is that will be 4 .42 times 10 to the 5 joules.

03:19 Okay.

03:22 So this is the work done by weight or gravity.

03:27 And this is the work done by friction.

03:31 So my net work done, work net equals 6 .8 times 10 to the 4 joules.

03:43 Now we want to use the work energy theorem to find the velocity of the elevator card just before it hits the spring.

03:53 Okay, and what the work energy theorem says is that the change in kinetic energy of an object equals the work done...

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