A chain 64 meters long whose mass is 20 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 3 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared. Your answer must include the correct units.
A chain 64 meters long whose mass is 20 kilograms is hanging over the edge of a tall building...
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1If the chain is uniform (mass is uniformly distributed), and is at rest at the beginning and the end of the operation, and there is no friction, then a simple computation of the potential energy before and after will do the trick, not forgetting the 3m at the top of the tower at the end. – 2012-09-29
1 Answers
Let $x$ denote the position along the chain in meters, with $x=0$ corresponding to the top of the chain at the top of the building. Let $\delta x$ denote a small length of chain. To figure out the total work, we break up the chain into small pieces of equal length $\delta x,$ compute the work to move each piece, and add it all together.
The top piece does not need to be lifted further. A piece at approximately $x$ meters down the chain needs to move $x$ meters if $x\le 3,$ or simply moves $3$ meters if $x\ge 3.$
The work required to lift a piece is $xF$ if $x\le 3,$ or $3F$ otherwise, where $F$ is the force due to gravity. Assuming the chain has constant density, the force acting on a small $\delta x$ of chain is $F=mg=(\frac{20}{64}\delta x)(9.8) = 3.0625 \delta x.$
So to move a small $\delta x$-length of chain takes $3.0625x\delta x$ Joules of work if $0\le x\le 3.$ Summing up the work over these small lengths corresponds to integrating $\int_0^3 3.0625xdx.$
The work required to move the rest of the chain is easier, since all the remaining $\delta x$-lengths move the same distance: $3$ meters. So we can compte this all at once as $3F=3mg=3(\frac{20}{64}61)(9.8)=560.4375$ Joules of work.
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0Ryan: Simple correction would suggest to include in your question the indications and/or full answer you received from the online WebWork. – 2013-01-06