Question

When the 55-kg cylinder is released from rest, the spring is subjected to a tension of 75 N . Deter mine the speed of the cylinder after it has fallen 200 mile meter. Express your answer with the appropriate units. V = How far has it fallen when it momentarily stops? (Figure 1) Express your answer with the appropriate units.

Answer 1

0.5mv^2 - mgh + 2T times h + 0.5 times k times s_p = 0 0.5 times 55 times v^2 - 55 times 9.8 times 0.2 + 2 times 75 times 0.2 + 0.5 times 300 times (0.4)^2 = 0 27.5v^2 - 107.8 + 30 + 24 = 0 27.5v^2 = 53.8 v = 1.95 m/s b)here let x be the distance where it stops 0.5mv^2 - mgh + 3T times h + 0.5 times k times s_p = 0 0 - 55 times 9.8 times x times 2 times 75x + 0.5 times 300 times (2x)^2 = 0 - 539x + 150x + 600x^2 = 0 x = 0.648 m