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Section 6-6 : Work

3. A cable with mass \(\frac{1}{2}\) kg/meter is lifting a load of 150 kg that is initially at the bottom of a 50 meter shaft. How much work is required to lift the load ¼ of the way up the shaft?

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Hint : What is the total mass of the chain and load at any point in the shaft? How does that relate to the force required to hold the chain and load at any point in the shaft?
Start Solution

Let’s start off with the convention that \(x = 0\) defines the bottom of the shaft and \(x = 50\) defines the top of the shaft. Therefore, \(x\) represents the distance that the load has been lifted. After lifting the load by \(x\) meters there will be \(50 - x\) meters of the chain left in the shaft that needs to be lifted along with the load.

Therefore, after lifting the load \(x\) meters, the total mass of the chain left in the shaft as well as the load is,

\[\frac{1}{2}\left( {50 - x} \right) + 150\,{\mbox{kg}} = 175 - \frac{1}{2}x\,{\mbox{kg}}\]

We know that the force required to hold the chain and load at any point is just the total weight of the chain and load at that point. We also know that (because we are in the metric system) the weight of a given mass (in kg) is just then,

\[{\mbox{Weight}} = {\mbox{mass}} \times 9.8\]

where 9.8 is the gravitational acceleration.

The force required to hold the chain and load a distance of \(x\) meters above the bottom is then,

\[F\left( x \right) = \left( {9.8} \right)\left( {175 - \frac{1}{2}x} \right) = 1715 - 4.9x\] Show Step 2

For the limits of the integral we can see that we start with the chain and load at the bottom of the shaft (i.e. at \(x = 0\)) and stop ¼ of the way up the shaft (i.e. at \(x = 12.5\)). These values are then the limits for the integral.

The work is then,

\[W = \int_{0}^{{12.5}}{{1715 - 4.9x\,dx}} = \left. {\left( {1715x - 2.45{x^2}} \right)} \right|_0^{12.5} = \require{bbox} \bbox[2pt,border:1px solid black]{{21,054.6875J}}\]