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Use Hooke's Law (F = -k s), where:

- F is the spring's restoring force.
- k is the spring constant.
- s is the stretch.

The negative sign indicates the spring's restoring force is opposite to the direction of the stretch.

You have a plot of weight [N] versus stretch [m]. The data forms a linear trend: \(y = 3.662x + 1.67\).

Question: How much will the spring stretch if 51.7 grams is hung on the spring?

- Answer in centimeters with three significant figures.
- Write "N/A" if not enough information is given to answer.

Note: When you calculate your answer, do not use the negative sign in Hooke's Law. The negative sign simply indicates that the force direction is opposite to the stretch (or compression).

Answer :

Final answer:

The spring will stretch by 13.79 cm when 51.7 grams is hung on it.

Explanation:

To find the stretch of the spring when a weight of 51.7 grams is hung on it, we will use Hooke's Law. First, we need to convert the weight to Newtons. Since 1 g is equal to 0.0098 N, the weight in Newtons is 51.7 grams * 0.0098 N/g = 0.50546 N. Now we can rearrange Hooke's Law equation, F = -k * s, to solve for s, the stretch of the spring. Plugging in the values, we get 0.50546 N = -k * s. Rearranging further, we have s = 0.50546 N / -3.662 = -0.1379 m. Since the question asks for the answer in centimeters, we can convert -0.1379 m to centimeters by multiplying by 100. Therefore, the spring will stretch by 13.79 cm when 51.7 grams is hung on it.

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