Darth Maul has taken a vertical spring (ignore the mass of the spring itself) wi
ID: 1785176 • Letter: D
Question
Darth Maul has taken a vertical spring (ignore the mass of the spring itself) with a spring constant of 810 N/m and has attached it to a table. The spring is then compressed down a distance of 0.100 m. Note: Be VERY careful to include all PE and KE contributions when solving this problem! (a) If Darth Maul places his light saber (mass 0.500 kg) on the compressed spring, with what upward speed can the spring give to the light saber when released? 05 (b) How high above its original position (spring compressed) will the light saber fly? x m/sExplanation / Answer
potential energy = mgh
= 0.500*9.81*0.100
= 0.4905 J
The initial energy in the spring was ½kx² = ½*810*0.100² = 4.05 J
So the initial kinetic energy of the ball K = 4.05 - 0.4905 = 3.5595 J
Ball's intital velocity =(2K/m) = [2*3.5595/0.5]
= 3.8 m/s
__________________
Ball will rise a further height h such that kinetic energy lost = potential energy gained:
½mv² = mgh
h = v²/(2g) = 3.77²/(2*9.81)
= 0.72m
Total height gained = 0.1 + 0.72 = 0.82 m
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