# Spring launched box sliding over friction surface

1. Sep 27, 2014

### Yae Miteo

1. The problem statement, all variables and given/known data

A block with mass m = 14 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4085 N/m after being compressed a distance x_1 = 0.546 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.6 m long. For this rough path, the coefficient of friction is μ = 0.43.

(This is one part of a multi-part problem. The current part is part 5.)

Instead, the spring is only stretched a distance x_2 = 0.162 m before being released.
How far into the rough patch does the block slide before coming to rest?

2. Relevant equations

$$F=-\int kxdx$$

$$v^2 = v_o^2 + 2ax$$

$$F=ma$$

3. The attempt at a solution

using
$$v^2 = v_o^2 + 2ax$$
solve for x
$$x = -\cfrac{mv_o^2}{2(F-f)}$$
plug in F
$$x = \cfrac {mv_o^2}{2(k \int xdx - mg \mu)}$$
but I'm not sure where to go from here.

2. Sep 27, 2014

### Satvik Pandey

I think you should use conservation of energy.
Firstly potential energy of spring will get converted into kinetic energy.From this you can find initial velocity of the block (velocity after leaving the spring).Then use equations of motion to find the answer.
After leaving the spring what will be the acceleration of the block?

3. Sep 27, 2014

### Simon Bridge

Presumably the rough patch lies beyond the maximum extension of the spring+length of the box.
To proceed - you should explicitly state your reasoning: what is your strategy?
i.e. where does the force F come from? If it comes from the spring: consider - is the spring still pushing on the crate when there is friction?

Have you tried using a work-energy type argument?
(Satvik is suggesting the long way around ... where does the energy of the spring finally end up?)