Question

A 10 ft. pole weighs 30 lbs. (of force, to be clear). It lies flat on the ground. You lift one end of the pole until it stands vertically, the other end of the pole has been braced to the ground while you lift the other end up. Once the pole is upright, the segment of length Δx at height x has been raised a vertical distance of x feet. How much work is done to raise the pole vertically?

Answer 1

Answer:

work done to raise the pole in vertical position is 150 lb ft

Explanation:

As we know that change in the potential energy of the pole is equal to the work done to raise the pole into vertical position

So here we can say that potential energy of small element of pole which is x height above the ground is given as

$dU = (dm)g x$

here we know that

$dm = \frac{m}{L}dx$

now we have

$dU = \frac{m}{L}dx g x$

now total potential energy of the pole is given as

$U = \int \frac{mg xdx}{L}$

so we have

$U = \frac{mgL}{2}$

here we know that

$U = \frac{30 \times 10}{2}$

$U = 150 lb-ft$

SO work done to raise the pole in vertical position is 150 lb ft

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Topic : Gravitational Potential energy

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