6.4 Work Question #3 (Medium)
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6.4 Integral Applications
Force and Work Application
Question #3 (Medium): Work Done to Vertically Lift an Object
It is important to distinguish whether the units are in the International or the US metric system. Pound (lb) in the
US system already includes the gravity, so it is not multiplied by . It is directly used as force.
Using integral to find the work done to vertically move an object, -axis can be used to represent the vertical
line along which the object moves. So any part of the segment can be considered as . Then the total weight of
the object is converted into the unit mass of
. Once the force is expressed using the unit mass, that unit
force is multiplied by the distance the segment travels, and that expression is integrated over the whole interval
to calculate the total work done:
cement package is lifted from the ground to the top of a building that is high. The rope that pulls the
metal tray containing the cement package weighs . The metal tray weighs . Calculate the total
Cement package and the metal tray weigh in total: .
The unit is in , so gravity is not included. So,
the cement and the tray. Then, .
For any part of the rope of length , the mass is . At height of , that part of the rope travels
meters to the top of the building. Part of the rope already at the top of the building travels meters. Part of the
rope touching the ground travels meters. So the interval is .
For any part of the rope, the force for is then:
. Then adding the work done to lift the cement package and
the metal tray, total work is: .
Therefore, total work required to lift the cement package using metal tray to the top of a building is .
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