A 2.00-kg package is released on a 53.1degree incline. 4.00 m from a long spring

Question

A 2.00-kg package is released on a 53.1degree incline. 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline (Figure 1). The coefficients of friction between the package and the incline are mu_s = 0.40 and mu_k = 0.20 the mass of the spring is negligible What is the speed of the package just before it reaches the spring? What is the maximum compression of the spring? the package rebounds back up the incline. How close does it get to its initial position?

Explanation / Answer

: 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

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