Ask a New Question

Question

A spring with a spring constant k=450.0 N/m is attached to a mass of m=8 kg and stretched -0.5 m from its equilibrium position and released.

What is the speed of the block when it is at the equilibrium position?
12 years ago

Answers

Jennifer
When it has stretched from its maximum position, the potential energy stored in the spring is 1/2*k*x^2 = 1/2*450*.5^2 = 56.25

this is converted to kinetic energy:

56.25 = 1/2*m*v^2 = 1/2*8*v^2

Solve for v
12 years ago

Related Questions

A spring has a spring constant of 259N/m. Find the magnitude of the force needed (a) to stretch the... A spring with a spring constant k of 44 N/m is stretched a distance of 30 cm (0.3 m) from its origin... A spring with a spring constant of 1.50 102 N/m is attached to a 1.2 kg mass and then set in motion... A spring with a spring constant of 223.9 N/m is compressed by 0.220 m. Then a steel ball bearing of... A spring has a spring constant of 42 N/m. How much work is required to stretch the spring 1.5 cm fro... 2. A spring of spring constant 220 N/m is sitting on a table. You pick it up and stretch it so that... A spring with a spring constant of 10 N/m is stretched 0.25 meters and released. How much force is n... A spring of spring constant 30.0 N/m is attached to different masses, and the system is set in mo... A spring has spring constant 0.3 m/newton. What force is necessary to stretch the spring by 1.5 mete...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use