- #1

Eduardo Leon

- 6

- 0

## Homework Statement

Whats is the final speed of the second box(mass = 1kg) when the first one (mass = 5 kg) has descended a distance equal to 0.6 meters in the rough ramp inclined 60° respect the x axis. Consider the spring is compressed a length x = 0.2 meters. The second box is tied to the spring and the system is released from its stationary state. For this case, consider the rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted), the dynamic friction coefficent is 0.2, and the elasticity constant of the spring k=200 Newtons/meters?

- Second box(mass = 1kg)
- First box(mass = 5 kg)
- Distance descended to 0.6 meters (rough ramp inclined 60° respect the x axis)
- The spring is compressed a length x = 0.2 meters.
- The rope, the spring and the pulley are ideal (have no mass and the rope doesn't get elonged or contracted).
- Dynamic friction coefficent is 0.2
- Elasticity constant of the spring k=200 Newtons/meters

It's the 27^{th}excersice

## Homework Equations

- Free-body diagrams
- Work (W=Fxd)
- Mechanic Energy Balance (conservative and non conservative system)
- Cinematic equation for final speed V^2
_{Final}= V^2_{initial}+2 a(X_{final}- X_{initial})

V^2

_{Final}the square power of the speed at a

**t**time

V^2

_{initial}the square power of the speed at a initial time

a the acceleration

(X

_{final}the displacement at a the

**t**time

X

_{initial}the displacement at a the initial time

## The Attempt at a Solution

The speed I calculated is equal to 3.68 m/s

The acceleration of the system when released (thanks to the free-body diagrams) = 11,29m/s^2

In order to find the speed of the second box, I first found the height where it will arrive after the spring releases, using energy balances. Then, after findig the acceleration via Free-body diagram used cinematic equations for speed according to de enunciated distance.