There are a number of distances than need to be defined. The slope distance, s1, is still the same, but lets define s2 as the horizontal distance from the slope to the uncompressed spring. Next, lets set s3 to be the rebound distance from the spring and ss be the compressed spring distance. All these distances are shown on the diagram at the left.
The distance ss must be calculated before the rebound distance s3 can be determined.
Part 1): Use the Principal of Work and Energy over distances s1, s2, and ss:
T1 + ΣU1-2 =
0 + Σ(Ug + Us + Uf.) = T2
mgh - ½kss2 - (μN1s1 + μN2s2 + μN2ss) = 0
where v1 = 0 v2
s1 = 10/sin26.6 = 22.33 ft
s2 = 6 ft
N1 = mg cos26.6 = 44.71
N2 = mg = 50 lb
Substituting the known values gives
5 (10) - 0.5 (50) ss2 - 0.3 (44.71) (22.33)
- 0.3 (50) (6) -
0.3 (50) ss = 0
110.5 - 25 ss2 - 15 ss = 0
Solving gives the solution for how far the spring is depressed before the package comes to rest,
ss = 1.824 ft (other root is negative, -2.424 ft)
Part B): Now that spring compressed distance is known, the total distance the package rebounds can be determined. The energy equation can be used again, but analysis is from the compressed position (v = 0) and the final position (v = 0). There are two distances involved,
ss and s3.
Σ(Us + Uf) = 0 (since comes to rest)
½kss2 - μN2s3 - μN2ss = 0
50 (1.82)2 - mg (0.3) s3 - mg (0.3) 1.82 = 0
s3 = 3.70 ft (or x = 6 - 3.7 = 2.4 ft)
Notice, the weight (W = mg) does not cancel and thus the weight does effect the final solution, even though it will be small if the spring compression distance is small.