Now assume that the water is contained in a 0.1-kg aluminum pot (cAl = 900 J/kg-K) that is initially at 293 K just like the wate
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From the given information in the question, the correct option is Option 1: 14 cm.
A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:
PE = k
Where PE is the energy, k is the spring constant and x is extension.
i. Given that: PE = 10 J and x = 10 cm, then;
PE = k
10 = k
20 = 100k
k = 0.2 J/cm
ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.
PE = k
20 = (0.2)
40 = 0.2
= 200
x =
= 14.1421
x = 14.14 cm
So that;
x is approximately 14.00 cm.
Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.
For more information, check at: brainly.com/question/1352053.