## DanDanDan 4 years ago 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

1. IsTim

W=E?

2. anonymous

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

3. Dandandan

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

4. IsTim

Yup.

5. Dandandan

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

6. IsTim

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

7. Dandandan

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?

8. AravindG

hey salini1!

9. anonymous

|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