document.write( "Question 158168: Hooke's Law states that the distance, d, a string is stretched varies directly with the force, F, applied to the spring. If a force of 12 pounds stretches a spring 3 inches, how far will the spring stretch when a force of 30 pounds is applied? \n" ); document.write( "
Algebra.Com's Answer #116540 by jim_thompson5910(35256)\"\" \"About 
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Since the \"distance, d, a string is stretched varies directly with the force, F\", this means that we have the equation \"d=kF\"\r
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\n" ); document.write( "\n" ); document.write( "\"d=kF\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"3=k%2812%29\" Plug in \"d=3\" and \"F=12\"\r
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\n" ); document.write( "\n" ); document.write( "\"3%2F12=k\" Divide both sides by 12 to isolate k\r
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\n" ); document.write( "\n" ); document.write( "\"1%2F4=k\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "So the variation constant is \"k=1%2F4\" (or in decimal form \"k=0.25\")\r
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\n" ); document.write( "\n" ); document.write( "\"d=kF\" Go back to the original equation\r
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\n" ); document.write( "\n" ); document.write( "\"d=%281%2F4%29F\" Plug in \"k=1%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "\"d=%281%2F4%29%2830%29\" Since we want the distance when the \"force of 30 pounds is applied\", we need to plug in \"F=30\"\r
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\n" ); document.write( "\n" ); document.write( "\"d=30%2F4\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"d=15%2F2\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "\"d=7.5\" Convert to decimal form\r
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\n" ); document.write( "\n" ); document.write( "So the distance is 7.5 inches when a force of 30 pounds is applied.\r
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