A force of 192 lb is required to hold a spring stretched 3 ft beyond its natural length. How much work W is done in stretching it from its natural length to 6 inches beyond its natural length

Answers

Answer 1

Answer:

The work done in stretching the spring is 8 lb.ft

Step-by-step explanation:

Given;

Applied force, F = 192 lb

extension of the spring, x = 3 ft

Determine the spring constant from the applied force and extension;

[tex]k = \frac{F}{x} \\\\k = \frac{192 \ lb}{3 \ ft} \\\\k = 64 \ lb/ft[/tex]

When the spring is stretched 6 inches beyond its natural length, the work done is calculated as follows;

x = 6 inches = 0.5 ft

[tex]W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} (64 \ lb/ft)(0.5 \ ft)^2\\\\W = 8 \ lb.ft[/tex]

Therefore, the work done in stretching the spring is 8 lb.ft


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Ok so do 9 times itself 5 times so basically its this:

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Next do the same but 3 times:

9*9*9

Exponents are not this:

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Another way to say it:

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There is your answer hope this helps

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Step-by-step explanation:

Ok, this is how you find exponents:

Well, exponents are really just abbreviated multiplication, so to find the numerical value of 9^5 and 9^3, you need to multiply 9 by itself 5 times and 3 times, then divide.

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Then you divide: 59049/729= 81.

But the question asks to rewrite the answer using a single positive exponent, so now we simplify 81 into an exponent.

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Another simple trick for solving exponents is to subtract the exponents:

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I hope this answer was helpful!

Giveaway Contest!
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Answers

Jelly sells 62 shares of stock she owns for a total of $433. If the stock was in two different companies, one selling at $6.50 a share & the other at $7.25 a share, how many of each did she sell?
----
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---------------------------
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-----------------------------
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