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).

Question

Potter

Physics

12 October, 04:42

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).

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Answers (1)

Know the Answer?

Teagan Cardenas · Answer 1

x = 0.138 m

Explanation:

In this exercise they give us the equation and the line that corresponds to Hooke's law, write the two equations and match the terms

F = k Δx = k (x - x₀)

y = 3.662 x + 1.67

Like terms

F = y

kx = 3.662 x

Since the graph is N vs m, the units of the constant are

k = 3.662 N / m

k x₀ = 1.67

Where

x₀ = 1.67 / k

x₀ = 1.67 / 3.662

x₀ = 0.456 m

This value is the natural length of the spring

We write Hooke's law with the mass applied

W = k x

mg = kx

x = mg / k

x = 51.7 10-3 9.8 / 3.662

x = 0.138 m