A vertical spring with a force constant of 700 N/m is initially at equilibrium. A 10kg mass is attached to the end of the spring, causes it to elongate. What will be the maximum elongation/deformation of the spring?

use work-energy theorem

Question

A vertical spring with a force constant of 700 N/m is initially at equilibrium. A 10kg mass is attached to the end of the spring, causes it to elongate. What will be the maximum elongation/deformation of the spring?

use work-energy theorem

istim · Answer

W=E?

anonymous · Answer

Assuming this is a vertical arrangement. $$W_s = {1 \over 2} k x^2$$$$E_m = mgx$$

dandandan · Answer

so $$(1/2) kx^2=mgx ?! $$

istim · Answer

Yup.

dandandan · Answer

i asked this earlier but that guys said ma=1/2*kx^2.. is it the same ?!

istim · Answer

I dunno. It doesn't look like it, but I recognize that configuration.

dandandan · Answer

A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m.  What is the coefficient of friction between the block and the tabletop?

aravindg · Answer

hey salini1!

anonymous · Answer

|dw:1328026340799:dw|
by work-energy theorem change in k.e=0 as when we consider that last point when spring has zero velocity,that is when it is pulled backward
so that is equal to Net WORK DONE=work done by spring force-mge
where e is the extension of spring that is haow
mg*e=1/2ke^2