A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential en

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case?

a.
b.
c.
d.

Oct 18th, 2017

First, let's remember the potential energy stored by the spring would be:

E = (1/2)(K)(X^2) ;   where: E is the potential energy (in Joules), K is the constant of the spring (in N/m) and X is the distance (in this case what the spring is compressed, in meters).

So from the above formula we solve for K like this:

E = (1/2)(K)(X^2)  ----------->  2*E = 2*(1/2)(K)(X^2)  ------------> 2E = (K)(X^2)

2E = (K)(X^2)  ---------------> 2E/X^2 = (K)(X^2)/X^2  ---------------> 2E/X^2 = K    so    K = (2E)/X^2

So it is giving X = 12.0 cm (we need to convert it into meters) and W = 72.0 J. Then we have:

X = 12.0 cm * (1 m/ 100cm) = 0.12 m  ------> X = 0.12 m

Finally, we have:

K = (2(72.0 J))/(0.12 m)^2 = 144 J / 0.0144 m^2 = 144 N.m / 0.0144 m^2 = 10000 N/m

K = 10 000 N/m  , so option A

Jun 16th, 2015

...
Oct 18th, 2017
...
Oct 18th, 2017
Oct 19th, 2017
check_circle