Calculating work against gravity

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Homework Statement



Calculate the work against gravity required to build a structure out of brick. The structure has a square base with 10ft sides and is 20ft high. The density of brick is around 80lb/ft^3

Homework Equations



W=∫F*d

The Attempt at a Solution



First I found the volume where V=(102) Δy
Then I found the mass=(ρ)(V)=(80)(102) Δy
Then I found the force=(gravity)(m)=(32.2)(80)(102) Δy
Then I integrated between 0 and 20 with respect to y=∫(32.2)(80)(102)y dy
which gave me 128800y2 between 0 and 20= 51520000 ft*lb
my answer is not correct and I'm not sure if I'm computing properly.
 
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getty102 said:

Homework Statement



Calculate the work against gravity required to build a structure out of brick. The structure has a square base with 10ft sides and is 20ft high. The density of brick is around 80lb/ft^3

Homework Equations



W=∫F*d

The Attempt at a Solution



First I found the volume where V=(102) Δy
Then I found the mass=(ρ)(V)=(80)(102) Δy
Then I found the force=(gravity)(m)=(32.2)(80)(102) Δy
Then I integrated between 0 and 20 with respect to y=∫(32.2)(80)(102)y dy
which gave me 128800y2 between 0 and 20= 51520000 ft*lb
my answer is not correct and I'm not sure if I'm computing properly.

Tell us what you think the correct answer is.

RGV
 
My final answer came out to be 5.152*10^7 ft*lb, but that was incorrect.
 
getty102 said:
My final answer came out to be 5.152*10^7 ft*lb, but that was incorrect.

Yes, you said that. I was asking what you thought was the _correct_ answer.

RGV
 
According to my textbook, the answer is W=1.6*10^6 ft*lb.
 
getty102 said:
According to my textbook, the answer is W=1.6*10^6 ft*lb.

(1) Why did you multiply by g (the acceleration of gravity)? A brick's weight is _already_ the force exerted by gravity: the "pound" is a force unit. (That is what happens when people use the confusing and outdated Imperial system of measurements in *technical* discussions; errors creep in all too easily. Imperial units are OK at cocktail parties and in backyard chats, but they ought to be eliminated from scientific computations.)

(2) Some bricks are lifted different heights. The bricks on the bottom are not lifted at all; the ones at the top are lifted the full 20 ft. Others are lifted only part way up.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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