Leaky cement bucket A 350 kg-bucket containing 4650 kg of cement is resting on the ground when a

Leaky cement bucket A 350 kg-bucket containing 4650 kg of...

Leaky cement bucket A 350 kg-bucket containing $4650 \mathrm{kg}$ of cement is resting on the ground when a crane begins lifting it at a constant rate of $5 \mathrm{m} / \mathrm{min.}$ As the crane raises the bucket, cement leaks out of the bucket at a constant rate of $100 \mathrm{kg} / \mathrm{min.}$ How much work is required to lift the bucket a distance of $30 \mathrm{m}$ if we ignore the weight of the crane cable attached to the bucket?

Key Concepts

Related Rates

Related rates involve understanding how different quantities change with respect to one another. In contexts where an object's position, mass, or other attributes are changing simultaneously, it is important to relate the rate of change of one quantity (like mass leakage rate) to another (such as the lifting speed). This connection enables the formulation of functions that accurately describe the system's behavior over time or distance.

Work

Work in physics is defined as the energy transferred by a force acting over a distance. When the force is constant, work is simply the product of the force and the distance over which it is applied. However, in scenarios where the force varies, work must be determined by integrating the force over the path of motion.

Integration of a Variable Force

Integrating a variable force involves setting up an integral where the force is expressed as a function of the distance or time parameter. This integral sums up the infinitesimal amounts of work done over small segments, providing the total work done for the entire process of lifting or moving an object when the force is not constant.

Variable Mass Systems

In some physical problems, the mass of an object is not constant but changes over time or distance. This is significant because the weight, and thus the force due to gravity, depends on the mass. Evaluating work in such cases requires expressing the mass as a function of the relevant variable (such as time or position) and incorporating this into the integration process.

Transcript

00:01 Okay, so there was a question earlier on in our homework problems that is similar to this, and they look a bit intimidating, but you will see soon enough this question can be broken down into a much simpler form.

00:19 So we know that work is equal to the integral of force, and in this case, since we're dealing with time, we could say f of t and we could rearrange this since force equals mass times gravity in our case or mass times acceleration but our acceleration in this case is gravity since it's a bucket dropping to the ground we could rearrange this integral as m of t multiplied by g and that's equal to work and now the mass of the bucket's going to change over time.

01:06 So the numbers we have is that it's 350.

01:12 Well, i should say the mass of the bucket's changing over time because the water is going to be, or the cement is going to be leaking out.

01:24 And thus, we're starting out 350 plus 4 ,650.

01:36 And then we're changing at a constant rate of a hundred.

01:46 It's leaking out at a constant rate of 100t.

01:54 Since we're doing with a change in or similar to a slope, you could have that 100t.

02:03 And rearranging this, we have 5 ,000 minus 100t.

02:09 And you can see this is basically in linear form.

02:17 We have our y intercept and we have our rate or our m and this is our x and this is our b.

02:27 Thus, not too complicated.

02:30 We could plug this into our integral and our boundaries would be, in this case, they would be 0 to 6.

02:48 And how did we get that? well, it takes five minutes per meter, and we are going to 30 meters...

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