OpenStudy (anonymous):

A spring with a force constant 400 N/m hangs vertically. You attach a 2.2 kg block to it and allow the mass to fall. What is the max distance the block will fall before it begins moving upward?

5 years ago
OpenStudy (anonymous):

The concept: the change in potential energy of the block is EXACTLY equal to the elastic potential energy of the spring. Does that help?

5 years ago
OpenStudy (anonymous):

ya a little

5 years ago
OpenStudy (anonymous):

i'll get 21.5x = 405x^2 know what?

5 years ago
OpenStudy (anonymous):

elastic potential energy is give by 1/2 kx^2. So how can you get 405x^2 ?

5 years ago
OpenStudy (anonymous):

*given

5 years ago
OpenStudy (anonymous):

* 200 x^2

5 years ago
OpenStudy (anonymous):

there you go. So now, all you have to do is... simplify.

5 years ago
OpenStudy (anonymous):

that's the part i have trouble with

5 years ago
OpenStudy (anonymous):

Just cancel 1 'x' on both sides.

5 years ago
OpenStudy (anonymous):

oh ya

5 years ago
OpenStudy (anonymous):

so it would be 11 cm?

5 years ago
OpenStudy (anonymous):

yep. 10.75 cm.

5 years ago
OpenStudy (anonymous):

ok thanks a lot

5 years ago
OpenStudy (anonymous):

:)

5 years ago
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