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glossolalia
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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!