Posted by **Anonymous** on Saturday, March 31, 2012 at 3:45am.

A particle of mass m kg is attached to one end A of a model spring OA of natural length L=0.81m and stiffness K N/m. The spring is hung vertically with the end O fixed and end A attached to the particle, which moves in a vertical line. The particle's mass, stiffness and natural length of the spring are related to the expression KL=2mg, where g is the magnitude of the acceleration due to gravity. It is observed that when the particle is at its highest point during the motion, its distance below O is 1/2L.

Using the law of conservation of mechanical energy, find the distance, in m, of the particle below O when it is at its lowest point during the motion.

thanks!

## Answer this Question

## Related Questions

- physics - 1)An elastic material has a length of 36cm when a load of 40N is hung ...
- Physics - A particle P of mass m kg is attached to two fixed points A and B by ...
- Physics - Hung vertically, a mass less spring extends by 3.00 cm a mass of 739. ...
- Physics - A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left ...
- physics please help!! - Hung vertically, a massless spring extends by 3.00 cm ...
- physic - Hooke's law describes a certain light spring of unstretched length 30.0...
- physics - A spring (k = 400 N/m) is hung vertically. a) If a 5 kg mass is ...
- PHYSICS!!! - Hooke's law describes a certain light spring of unstressed length ...
- Physics - A spring with an unstrained length of 0.075 m and a spring constant ...
- Physics - A spring with an unstrained length of 0.066 m and a spring constant of...

More Related Questions