Question

a force of 100 n stretched a spring by 0.6 centimeters. if the spring is then stretched to 0.2 cm, what happened to the force applied to the spring? the amount of force applied increased to three times the original force. the amount of force applied was removed from the spring. the amount of force applied decreased to one - third of the original force. the amount of force applied decreased to one - third of the original force. the amount of force applied remained the same as the original force.

Explanation:

Step1: Recall Hooke's Law

Hooke's Law is $F = kx$, where $F$ is the force, $k$ is the spring - constant, and $x$ is the displacement. First, find the spring - constant $k$. Given $F_1=100\ N$ and $x_1 = 0.6\ cm=0.006\ m$. From $F = kx$, we can solve for $k$: $k=\frac{F_1}{x_1}=\frac{100}{0.006}=\frac{100000}{6}\ N/m$.

Step2: Calculate the new force

Now, $x_2 = 0.2\ cm = 0.002\ m$. Using $F = kx$ again with the same $k$, we have $F_2=k x_2$. Substitute $k=\frac{100000}{6}\ N/m$ and $x_2 = 0.002\ m$ into the formula: $F_2=\frac{100000}{6}\times0.002=\frac{200}{6}=\frac{100}{3}\ N$.

Step3: Compare the new force with the original force

The original force $F_1 = 100\ N$, and the new force $F_2=\frac{100}{3}\ N$. So the new force is one - third of the original force.

Answer:

The amount of force applied decreased to one - third of the original force.