1. ## Another work problem

a 1600 lb elevator is suspended by a 200 ft cable that weighs 10 lb/ft. How much work is required to raise the elevator from the basement to the third floor, a distance of 30 feet?

I know work is just force times distance, and thus the distance would be 30. But I have NO IDEA what the force would be in this problem or how to find it. Could somebody help me out?

2. Okay, so:

$W = \int_{ {l }_{i } }^{ {l }_{ f} } Fdl$

But in this case F is not a constant.

Remember, F = ma = mg

But m is a function of the length of the cable here, so F is a function of the length of the cable as well.

What is the mass of the system in terms of the length of the cable?

mass = 1600 + 200*(10) - 10*(distance elevator has risen)

See how the mass decreases 10 lb for every foot the elevator has risen?

So then F = mg = (3600 - 10L)g, where L is the distance it has risen
Then we have:

$W = \int_{ 0 }^{ 30 } g(3600 - 10l)dl = 9.8\int_{ 0 }^{ 30 } (3600 - 10l)dl$

And solve that bad boy. As a check, it should be less than if the mass stayed constant which would be 9.8*30*(3600) = 1058400

3. Originally Posted by Mentia
Okay, so:

$W = \int_{ {l }_{i } }^{ {l }_{ f} } Fdl$

But in this case F is not a constant.

Remember, F = ma = mg

But m is a function of the length of the cable here, so F is a function of the length of the cable as well.

What is the mass of the system in terms of the length of the cable?

mass = 1600 + 200*(10) - 10*(distance elevator has risen)

See how the mass decreases 10 lb for every foot the elevator has risen?

So then F = mg = (3600 - 10L)g, where L is the distance it has risen
Then we have:

$W = \int_{ 0 }^{ 30 } g(3600 - 10l)dl = 9.8\int_{ 0 }^{ 30 } (3600 - 10l)dl$

And solve that bad boy. As a check, it should be less than if the mass stayed constant which would be 9.8*30*(3600) = 1058400
whoa, whoa, whoa, I have no idea what you're talking about, I don't think mass even plays a part in this problem. It shouldn't anyways since we haven't covered that yet.

4. Mass must play a part. The force on the system (the elevator and the cable) is equal to its mass times it acceleration. On earth, the acceleration due to gravity is 9.8 m/s.

And yes i appear to have made another mistake. 9.8 m/s is 32 ft/s. So instead of multiplying by 9.8 you should multiply by 32.

Do you see how the mass changes though? As the cable is rolled up at the top, the lifting mechanism doesnt need to pull as much weight and so the force it needs to provide decreases as the elevator goes up.

As a final answer I get 3312000 foot-pounds