A 2.34 kg package is released on a 50.7 degree incline, 4.00 m from a long sprin

Question

A 2.34 kg package is released on a 50.7 degree incline, 4.00 m from a long spring with force constant 139 n/m that is attached at the bottom of the incline . The coefficients of friction between the package and the incline are 0.36 and 0.2. The mass of the spring is negligible.


What is the speed of the package just before it reaches the spring? Note you must use the numbers given in the introduction for the mass and angle not from the figure.

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

m = 2.34kg, =50.7, s = 4m, k = 139, _s = 0.2, _k = 0.36

v = (2as) = (2*(gsin-(_k)gcos)*s) = (2*(7.58-2.234)*4) = 6.53m/s

speed of the package when it reaches spring = 6.53m/s

0.5*k*x^2 = 0.5*m*v^2 - (_k)mgcosx + mgxsin

0.5*139*x^2 = 0.5*2.34*6.53^2 - 5.22x + 17.74x

69.5x^2 - 12.52x - 49.88 = 0

x = 0.94m

maximum compression = 0.94m

mghsin + (_k)*mghcos = 0.5*k*x^2

2.34*9.8*(sin50.7)h + 0.36*2.34*9.8*(cos50.7)h = 0.5*139*0.94^2

h = 61.47/(17.74 + 5.22) = 2.677metres

Therefore distance from initial position = 4 - 2.677 + 0.94 = 2.26metres

0.94 has been added because h is calculated from the compressed point.

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