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Original Question
It takes 3.38 J of work to stretch a Hooke’s-law spring 5.63 cm from its unstressed length. How much the extra work is required...Asked by Maddy
It takes 3.78 J of work to stretch a Hooke’s-law
spring 14.2 cm from its unstressed length.
How much the extra work is required to
stretch it an additional 11.2 cm?
Answer in units of J.
spring 14.2 cm from its unstressed length.
How much the extra work is required to
stretch it an additional 11.2 cm?
Answer in units of J.
Answers
Answered by
Damon
Now this is so similar to your last question that I hesitate to say aanything.
remember from the other one:
potential energy stored = (1/2) k x^2
so
3.78 Joules = .5 * k * (.142)^2
solve for k
now TOTAL work to get to 14.2 + 11.2 or 25.4
= .5 * k * (.254)^2 Joules
now subtract 3.78 from that
remember from the other one:
potential energy stored = (1/2) k x^2
so
3.78 Joules = .5 * k * (.142)^2
solve for k
now TOTAL work to get to 14.2 + 11.2 or 25.4
= .5 * k * (.254)^2 Joules
now subtract 3.78 from that
Answered by
Maddy
Oh!! I get it, I just had trouble finding out how to solve to get k, but then I realized you just have to divide both sides until you isolate k by itself. Thanks!
Answered by
Damon
You are welcome.
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