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Sidana [21]
3 years ago
13

A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b

y 10 cm. How far would the spring need to be stretched to give the spring 20 J of elastic potential energy?
Use proportional reasoning.

Option 1: 14 cm
Option 2: 28cm
Option 3: 20 cm
Option 4: 40 cm
Physics
1 answer:
Firlakuza [10]3 years ago
8 0

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 = \frac{1}{2} k x^{2}

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 = \frac{1}{2} k x^{2}

10 = \frac{1}{2} k 10^{2}

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 = \frac{1}{2} k x^{2}

20 = \frac{1}{2} (0.2) x^{2}

40 = 0.2 x^{2}

x^{2} = 200

x = \sqrt{200}

  = 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.

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<u>Explanation:</u>

Given data

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It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

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