Question

1)    A 2.00-kg package is released on a 53.1

Answer 1

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