Answer to Question #128039 in Physics for Jackson

Question

A force of 6N extends a spring of natural length 0.6m by 0.02m .What will be the length of the spring when the applied force is 30N.

Answer 1

According to the Hooke's law, the force $F$ and extension $\Delta x$ are connected in a following manner:

$$F = k\Delta x$$

where $k$ is a spring constant. Expressing $k$ using the first measurement, get:

$$k = \dfrac{F_1}{\Delta x_1} = \dfrac{6N}{0.02m} = 300 N/m$$

Next, let's find the extension for the second force:

$$\Delta x_2 = \dfrac{F_2}{k} = \dfrac{30N}{300 N/m} = 0.1m$$

The total length is the sum of the natural length 0.6m and the extension:

$$l_2 = l_0 + \Delta x_2 = 0.6m + 0.1m = 0.7m$$

Answer. 0.7 m.