Incline + Spring = Hours of Fun. Sort of.

[glossolalia]

Homework Statement

A 1.71 kg package is released on a 54.1 deg incline, 4.00m from a long spring with force constant 141 N/m that is attached at the bottom of the incline . The coefficients of friction between the package and the incline are 0.39 and 0.24. The mass of the spring is negligible.

a) What is the speed of the package just before it reaches the spring?
b) What is the maximum compression of the spring?
c) The package rebounds back up the incline. How close does it get to its initial position?

Homework Equations

U_el = 1/2kx^2, U_grav = mgh
F=ma, etc. etc.
v^2 = v_0^2 + 2a∆x

The Attempt at a Solution

Got part a by finding out the force down the incline due to gravity less the force up the incline due to friction, solved for acceleration and plugged the value into v^2 = v_0^2 + 2a∆x to get 7.24 m/s, a correct answer. If there was an easier way, someone please point it out.

So, 1.71(9.8)(sin(54.1))-1.71(9.8)(cos(54.1))(.24) = 11.22N
11.22N/1.71kg = 6.56 m/s^2
sqrt (2*6.56m/s^2*4m) = 7.24m/s


part b, i quadratikifized it, letting x = the distance the spring compresses and using the formula U_grav - W_friction = 1/2kx^2 or 1.71(9.8)(4+x)(sin54.1) - .24(9.8)(1.71)(cos(54.1)) = 1/2(141)x^2, or 70.5x^2 -11.217x -44.867 = 0, yielding the result of .881m for x, also correct.

So now I've got this box at the bottom of a compressed spring. Its elastic energy is 1/2kx^2 or 54.7J. It's going to shoot back up, fighting friction (2.35N) and the x-component of gravity (13.57N), and that's where I get a bit stuck. I want to know when the only energy remaining is potential, I'm figuring. If I could figure out the velocity of the box as it left the spring, I'd be set, but I don't know how to solve for that.

Someone give me a nudge? Thankee!

 

[PhanthomJay]

Well, when solving for the distance the spring is compressed, your equation did not show any Kinetic Energy terms. Why not?

 

[SammyS]

The Work - Energy Theorem.

The block's change in Kinetic Energy is equal to the Net Work done on the block.

It might not get up the incline far enough to leave the spring.

 

[glossolalia]

Okay - still getting a wrong answer. It's because I'm approaching it the wrong way, I know, but wanted to clue you into my thought process here.

So, we have 54.7J of potential elastic energy sitting down there and we want to find out how high up that box gets when the elastic energy is transformed to kinetic energy, some of which is dissipated by friction, which is then converted completely into gravitational potential energy. So

Let mgh = U_grav, except in this case, let mg(sin(54.1))x = U_grav

So U_el - W_friction - W_grav = U_grav

or 54.7J - 2.36N(x meters) - 13.57N(x meters) = 16.75N(sin(54.1)(x meters) or
54.7J = 29.50N(x meters) -> x = 1.85m traveled. Since the origin of part C is 4.881m from the origin of part A, 4.881-1.85=3.03m. This answer is incorrect. What fundamental am I missing here?

Btw, Jay - thanks, there is no kinetic energy in my initial equation because there is no kinetic energy at either the initial or final position.

(Question specifically states "Here the initial position should be at the maximum spring compression and the final position should be at the final height. But be careful, it doesn't ask for how high you go, it asks for how far below your initial position you are."

 

[glossolalia]

Ok, ok, so I WASN'T thinking clearly. We want to know when the U_el EQUALS U_grav (less the energy lost to friction). I was sillily removing energy from the equation when I was subtracting the work done by gravity.

So, U_el = F_frict(x meters) + U_grav
or
54.7J = x(2.36N+13.57N) -> x = 3.434m -> subtract compressed length (.881m) and subtract that figure from 4m and we get 1.45m from the starting position of the block.

(Whew!)

Thanks, fellas!

 

[SammyS]

Good !