A 120m120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15kg15\operatorname{kg}/mm. How much work is done to pull 80m80\operatorname{m} of the chain to the top of the building?

940800 J ⁡ 940800\operatorname{J}

A 120 m 120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15 kg 15\operatorname{kg} / m m . How much work is done to pull 80 m 80\operatorname{m} of the chain to the top of the building?

A 120 meter chain hangs freely from the side of a building. The chain weighs 15 kilograms per meter. How much work is done to pull 80 meters of the chain to the top of the building? So in this problem, we have some kind of building that we are dealing with, And what we know about this building is that there's a chain that is freely dangling over the side of this building. Now we're told that this full chain here is dangling over this building and has a length of 120 meters. Now what we're trying to solve in this problem is the work required to lift 80 meters of the chain to the top of the building. So if we lift 80 meters up, we'll say, have about that much of the rope remaining right there. And that's after we've lifted the chain 80 meters up. So that is gonna go ahead and give us the remainder of chain that we have. And I'm not drawing this perfectly to scale here, but this just means that we've lifted a certain portion of chain up, and we wanna know how much work is required to lift this portion up. Well, we should know the general work equation. General work equation says that we need to take our force function and integrate it. Now in this specific situation, we're dealing with a lifting problem. And recall that there's actually another version of this specific equation we can use when it comes to solving lifting problems that looks something like this. The work is gonna be the integral from zero to h of our weight per length, this is like a weight density right here, multiplied by the length function, all integrated with respect to y. This is the equation we can use when it comes to solving these types of lifting problems. Now what I'm first going to do is figure out what my weight per length is. Now notice that we're told this value right here for the chained weight, but notice the units here are in kilograms per meter. This is not a weight density, but rather a mass density, so we have kilograms and meters. If I want to get this in a weight per length, I need to multiply this value by 9.8, Earth's gravitational acceleration, in meters per second squared. So we'll have 15 times 9.8, and this all comes out to 147. And now the weight here is gonna be in newtons, so we're gonna be dealing with newtons per meter. And that right there is going to be our weight density, either the weight per length. So we now have the weight per length figured out, and next we need to figure out what this length function is right here. Well, to find l of y, our length function, we can just do this. We can take our 120 meter distance right here, and we can subtract off y. Because recall that we're pulling more and more of this chain up the side of the building. As we pour pull more and more of this up, that means this length is gonna be getting smaller right here. So we're subtracting it by whatever distance y we're pulling this chain up. So this right here would be our function l of y. And so to put everything together, we're going to have that our work is gonna go from zero all the way up to the distance we're lifting it, which in this case is 80 meters, so from zero to 80. And then we're going to have our length function right here multiplied by the weight per length. The weight per length, I can see, is one forty seven. Then we're going to multiply that by our length function, one twenty minus y, and all of that is going to be integrated with respect to y. So this is what we have. Now from here, I can take this one forty seven and pull it outside of the integral. And now I can take the antiderivative. The antiderivative of one twenty is one twenty y. The antiderivative of y is just gonna be y squared over two using the power rule. So this is what we get, and this is all gonna be bounded from zero all the way up to 80. Now notice in this problem that zero, if we plug that in for y, that's just gonna set everything to zero. So we only need to worry about our upper bound of 80. So you can take 80 and plug it in for y. Now I trust that that's something you could do. You could plug 80 in for y squared here, so 80 squared, and then plug 80 in for y right there. You can multiply everything and subtract these numbers, and then multiply that by one forty seven. And if you do all this math, you crunch these numbers, you should get 940,800, and your work is gonna be measured in joules. So this is the work required to lift the chain 80 meters from its current position if it's a 120 meter long length, and that is the solution to this problem. So this is how you can solve these types of problems where we need to find the work required to lift an object like a chain where we don't have some constant force being lifted, but rather a force that is continuously changing because more and more chain is being pulled up. So hope you found this video helpful, and let's move on and get some more practice.

A 120 m ⁡ 120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15 kg ⁡ 15\operatorname{kg} / m m . How much work is done to pull 80 m ⁡ 80\operatorname{m} of the chain to the top of the building?

940800 J ⁡ 940800\operatorname{J}

9600 J ⁡ 9600\operatorname{J}

4800 J ⁡ 4800\operatorname{J}

470400 J ⁡ 470400\operatorname{J}

10. Physics Applications of Integrals

Work

Calculus

10. Physics Applications of Integrals

Work

Struggling with Calculus?

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Multiple Choice

A $120 m$ chain hangs freely from the side of a building. The chain weighs $15 kg$ / $m$ . How much work is done to pull $80 m$ of the chain to the top of the building?

$940800 J$

$9600 J$

$4800 J$

$470400 J$

Verified step by step guidance

Step 1: Understand the problem. Work is calculated as the integral of force over distance. Here, the force is due to the weight of the chain being lifted, and the distance is the height over which the chain is pulled.

Step 2: Define the variables. Let the chain's weight per meter be $ w = 15 \text{ kg/m} $. The gravitational force per meter is $ F = w \cdot g $, where $ g $ is the acceleration due to gravity ($ g \approx 9.8 \text{ m/s}^2 $).

Step 3: Set up the integral. The work done to lift a small segment of the chain, $ dx $, is $ dW = F \cdot dx $. Since the force varies with the length of the chain being lifted, the total work is $ W = \int_{0}^{80} F(x) \cdot dx $, where $ F(x) $ is the weight of the chain lifted at a given height $ x $.

Step 4: Express $ F(x) $. At any height $ x $, the weight of the chain being lifted is proportional to the length of the chain above $ x $. Thus, $ F(x) = w \cdot g \cdot (120 - x) $, where $ 120 - x $ represents the remaining chain length.

Step 5: Solve the integral. Substitute $ F(x) $ into the integral: $ W = \int_{0}^{80} w \cdot g \cdot (120 - x) \cdot dx $. Expand and integrate term by term to find the total work done. Remember to evaluate the definite integral from $ x = 0 $ to $ x = 80 $.

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