## Decart Group Title A 0.321-kg mass is attatched to a spring with force constant of 13.3 N/m. If the mass is displaced 0.256 m from the Equalibrium and released, what is its speed when it is 0.128 m from equalibrium? one year ago one year ago

1. raspberryjam Group Title

$\frac{1}{2}mv^2=\frac{1}{2}kx_{total}^2-\frac{1}{2}kx^2$ where k is the spring constant and $x_{total}$ is the total displacement from equilibrium. $x$ is the distance 0.128 m.

2. raspberryjam Group Title

Law of conservation of energy yo. Kinetic energy = (potential energy of spring at total displacement) - (potential energy of spring at whatever point you're trying to find the velocity at)

3. Decart Group Title

okay

4. Decart Group Title

i thought the potential would be at total displacement like mgh

5. raspberryjam Group Title

Well you're trying to find the velocity at a specific point so you have to subtract it from the potential energy of when the spring is fully stretched which gives you the amt of potential energy transformed into kinetic energy.

6. raspberryjam Group Title

Also you can't use mgh in this case because it's a spring-based problem...Hooke's law

7. Decart Group Title

do I convert the energy into velocity

8. Decart Group Title

I was saying mgh as an example

9. raspberryjam Group Title

No just solve for v in the equation above

10. raspberryjam Group Title

You can't just do total displacement because you're trying to find the velocity at a specific point which is when x=0.128 m. You aren't finding the displacement when x=0.256 m.

11. Decart Group Title

that worked! I do not quite understan how though.