Answer:
Maximum spring length = 0.94m
Explanation:
Newton's law of forces in the vertical direction is given as:
Ef = Ks S - mg = 0
Ks = mg/ s
Ks = mg/ (L1 -Lo)
Ks =( 0.3 × 9.8 ) / ( 0.77 - 0.59 )
Ks = 2.94 / 0.18
Ks = 16.33N/m
The initial state is one in which the mass is pulled down so that the length of spring is 0.84m and the mass has initial speed of 1.8m/s. The final state is that in which the minimum point reached and the maximum length attained is L3
Ei = Ef
1/2mVi^2 + 1/2 Ks(L2-Lo)^2 + mg(L3- L2) = 1/2Ks(L3 - Lo)^2.
1/2 Mvi + 1/2 Ks(L2 - Lo)^2 + mg(L3 - L2)^2 = 1/2Ks(L3^2 - 2LoL3 + Lo^2)
1/2 KsL3 - (mg + KsLo)L3 + [1/2KsLo^2 + mgL2 - 1/2Ks(L2 -Lo)^2- 1/2mVi^2] = 0
Substituting the values to solve for L3 using the quadratic formula
1/2(16.33)L3^2 - (0.3×9.8) + 16.33(0.59)L3 + 1/2(16.33×0.59^2) + (0.3×9.8×0.84)- 1/2(16.33(0.84-0.59)^3 - 1/2( 0.3× 1.8^2] = 0
8.165L3^2 - 2.94 + 9.63L3 + 2.84 + 2.47 - 0.51 - 0.486
8.165L3^2 + 9.63L3 + 1.37
L3 =[ 9.63 +- sqrt(9.63^2 - 4(8.165)(1.37) / (2× 8.165)]
L3 =[ 9.63 +- sqrt(92.74 - 44.74) /16.33]
L3 = 9.63 +- sqrt(47.996)/16.43
L3 = 9.63 +- 6.93/ 16.33
L3 = (9.63 - 6.93)/16.34
L3 = 2.7/16.33
L3 = 0.165 + 0.77
L3= 0.935m