Sagot :
Answer:
1.I'm assuming that you are asking what is the elastic potential energy stored in the spring at the position stretched by 16.5 cm...
Since you know the spring constant k, 144 N/m and the spring stretch from the equilibrium position x, is 16.5 cm, or 0.165, you find the spring's potential energy from the equation PE = 1/2 kx² which equals 1.96 Joules, or kg *m²/s²if you want SI units.
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Explanation: