A 2.00-kg package is released on a 53.1? incline, L = 3.99 m from a long spring

Question

A 2.00-kg package is released on a 53.1? incline, L = 3.99 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline (see figure below). The coefficients of friction between the package and the incline are ?s = 0.40 and ?k = 0.11. The mass of the spring is negligible.

(a) What is the speed of the package just before it reaches the spring? m/s

(b) What is the maximum compression of the spring? m

(c) The package rebounds back up the incline. How close does it get to its initial position? m

Explanation / Answer

Ok, here we just need to use the work - energy theorem : Remember : Initial energy - final energy = work made by friction Initial energy = potential energy = m*g*h Let's suppose the spring compresses = x meters L = 4 + x H = (4 + x)*sin(53.1) = 0.8*(4+x) Initial energy = 2*g*0.8*(4+x) g = 9.8 m/s^2 Final energy = Only elastic energy, because when the spring reaches its maximum compression the kinetic energy is zero. Final energy = 1/2*120*x^2 Initial energy - final energy = work done by friction work = 2*9.8*cos(53.1)*0.2*(4+x) 2*9.8*0.8*(4+x) - 60*x^2 = 0.6*2*9.8*(4+x)*0.2 3.92*(4+x) = 60x^2 4+x = 4.5x^2 4.5x^2 - x - 4 x = 1.06 meters So the maximum compression of the spring is 1.06 meters Now, just now we can obtain the speed of the package just before it reaches the spring, because we have the altitute : H = (4 + 0.53)*0.8 = 3.6 meters Initial energy = 2*3.6*9.8 Final energy = kinetic energy = 2*v^2 / 2 = v^2 Work done = 2*9.8*cos(53.1)*(4)*0.2 2*3.6*9.8 - v^2 = 2*9.8*cos(53.1)*4*0.2 70.56 - v^2 = 9.42 v = 7.6 m/s >>> That's the speed c) Here we just need to use the work - energy theorem The initial energy now is the elastic energy : Initial energy = 1/2*120*0.53^2 Final energy = Only potential energy = 2*9.8*h Work = 2*9.8*cos(53.1)*h*cscte(53.1) 16.8 - 19.6h = 14.7h h = 0.5 meters That's the new altitude Hope that helps

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