Question

Use Hooke's Law for this (F = - k s ): Where F is the spring's restoring force; k is the spring constant; and s is the stretch. The negative sign means the spring's restoring force is opposite the stretch direction. You have a plot from weight [N] versus stretch [m]. The data forms a linear trend y = 3.662 * x + 1.67. How much will the spring stretch if 51.7 grams is hung on the spring? Answer in centimeters with three significant figures or N/A if not enough information is given to answer. When you calculate your ansswer, don't use the negative sign in the Hooke's Law formula. Just know that the negative sign simply denotes the force direction is opposite the stretch (or compression).

Answer 1

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

Step-by-step 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.