Question

Hi, I have a little question about Potential Energy.

Answer 1

How much additional potential energy is stored in a
spring that has a spring constant of 16.5 N/m if the
spring starts 10.0 cm from the equilibrium position
and ends up 17.0 cm from the equilibrium position?

Answer 2

I was thinking to apply: \(\tt U=0.5k\cdot x^2\)
and the additional energy stored, refers
to stretching from 10 cm to 17 cm.

So, that would be the change in the potential energy:
\(\tt \Delta U=U_f-U_i\)
\(\tt \Delta U=0.5k\cdot x_1^{~~2}~-~0.5k\cdot x_o^{~~2}\)

I will write these lengths in meters,
\(\tt \Delta U=0.5(16)\cdot 0.17^{2}~-~0.5(16)\cdot 0.1^{2}\)

Which gives me:
\(\tt \Delta U=0.1512~J\)

Answer 3

Is this correct ?

Answer 4

@Michele_Laino , can you please take a look if you have some time?

Answer 5

what is the length of the spring when no force is acting on it?

Answer 6

That is not given in the problem, as far as I can read.

Answer 7

without any other information, I think it is correct

Answer 8

Yes, but the site (where I submitted the answer) tells me my answer is wrong:(

Answer 9

oh, hold, I think I entered `6 instead of 16.5 into wolf.

Answer 10

16 instead of 16.5

Answer 11

using the correct constant,
k=16.5 (not 16)

I get,
\(\Delta \)U = 0.155925 J

Answer 12

Yes, I got it:)

Answer 13

ok! :)