Final speed and compression in a kinetic-potential energy problem?

**erik-the-red** · Oct4-05, 05:17 PM

A 2.00-kg package is released on a $LaTeX Code: 53.1 ^\\circ$ incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are $LaTeX Code: \\mu_{s} \\;=\\; 0.40$ and $LaTeX Code: \\mu_{k} \\;=\\; 0.20.$ The mass of the spring is negligible.

1. What is the speed of the package just before it reaches the spring?

2. What is the maximum compression of the spring?

3. The package rebounds back up the incline. How close does it get to its initial position?

**erik-the-red** · Oct4-05, 07:10 PM

I forgot that I violated the sticky rule. Sorry!

I started off by breaking the force of weight into x and y components. I used the equation

. But,

. I'm looking for the final velocity, which can be found in the final kinetic energy.

My answer was 8.50 m / s, but that was incorrect.

For the second problem, I used $LaTeX Code: K\\Delta X=f_k+mgsin(\\Theta)$ . Solving for $LaTeX Code: \\Delta X$ , I got .151 m, which is also incorrect.

I don't know how to do part C.

**Doc Al** · Oct4-05, 07:34 PM

Originally Posted by erik-the-red

I started off by breaking the force of weight into x and y components. I used the equation . But, . I'm looking for the final velocity, which can be found in the final kinetic energy.

My answer was 8.50 m / s, but that was incorrect.

The idea of using energy methods is fine. Show exactly what you did.

For the second problem, I used $LaTeX Code: K\\Delta X=f_k+mgsin(\\Theta)$ . Solving for $LaTeX Code: \\Delta X$ , I got .151 m, which is also incorrect.

When the spring is maximally compressed the force on the package will not be zero. Once again, use energy methods.

**erik-the-red** · Oct4-05, 09:43 PM

For part a, I did

.

So,

LaTeX Code: (1/2)(2.00)(v_2)^2 + (2.00)(9.80)(-3.20) = -.47

m/s.

But, that is also not correct.

**Doc Al** · Oct4-05, 10:15 PM

Show how you calculated the work done against friction.

**erik-the-red** · Oct4-05, 10:28 PM

Yeah, something was SERIOUSLY not right with that work done by friction.

I ended up getting the correct answer. I must have punched in something wrong in my calculator.

Thanks!