Monday

September 22, 2014

September 22, 2014

Posted by **Jess** on Wednesday, March 23, 2011 at 11:11pm.

- physics -
**drwls**, Thursday, March 24, 2011 at 10:39amMax spring potential energy

= (1/2) k X^2 = (100/2)*(0.02)^2

= 0.2 Joules

When 0.01 m below the release point, the spring has 1/4 of its max potential energy, so 0.15 J is available for kinetic energy

(1/2)MV^2 = 0.15 J can be solved for the speed V. I get 0.548 m/s

It is permissible to neglect gravity in this problem when you regard kx as the force applied by the spring, including Mg, with x measured from the equilibrium position. g does not affect the answer or the period of vibration.

- physics -
**arwa**, Wednesday, April 20, 2011 at 2:54pmthe answer is 1.73 but i don't know how can every one help me ??

**Answer this Question**

**Related Questions**

physics - Janet wants to find the spring constant of a given spring, so she ...

Physics - A spring with an unstrained length of 0.066 m and a spring constant of...

physics - A spring with an unstrained length of 0.074 m and a spring constant of...

Physics - A spring with an unstrained length of 0.076 m and a spring constant of...

Physics #43 - A spring with an unstrained length of 0.076 m and a spring ...

PHYSICS - a 0.1 kg mass is suspended at rest from a spring near the Earth's ...

physics - a 0.1 kg mass is suspended at rest from a spring near the Earth's ...

physics - If a 0.1 kg mass is suspended at rest from a spring near the Earth's ...

physics - The drawing shows three identical springs hanging from the ceiling. ...

Physics PLEASE HELP! - A spring with an unstrained length of 0.076 m and a ...