Hooke's law states that the force needed to stretch a spring is proportional to the amount the spring is stretched. If $10.00 \mathrm{~N}$ stretches a certain spring $5.0 \mathrm{~cm}$ , how much will the spring be stretched by a force of $9.00 \mathrm{~N}$ ? $\newline$ A force of $9.00 \mathrm{~N}$ will stretch the spring by $\square$ centimetres. $\newline$ (Simplify your answer. Type an integer or a decimal.)

Full solution

Q. Hooke's law states that the force needed to stretch a spring is proportional to the amount the spring is stretched. If

10.00 \mathrm{~N}

stretches a certain spring

5.0 \mathrm{~cm}

, how much will the spring be stretched by a force of

9.00 \mathrm{~N}

\newline

A force of

9.00 \mathrm{~N}

will stretch the spring by

\square

centimetres.

\newline

(Simplify your answer. Type an integer or a decimal.)

Find Spring Constant: First, let's find the spring constant $k$ using Hooke's Law, which is $F = kx$ , where $F$ is the force and $x$ is the extension.
Calculate k: We know that $10.00\,\text{N}$ stretches the spring $5.0\,\text{cm}$ , so $k = \frac{F}{x} = \frac{10.00\,\text{N}}{5.0\,\text{cm}}$ .
Find Extension for $9.00\,\text{N}$ : Calculate $k$ : $k = \frac{10.00\,\text{N}}{5.0\,\text{cm}} = 2.00\,\text{N/cm}$ .
Rearrange Formula: Now, we need to find the extension $x$ when the force $F$ is $9.00\,\text{N}$ using the same formula: $F = kx$ .
Plug in Values: Rearrange the formula to solve for $x$ : $x = \frac{F}{k}$ .
Calculate $x$ : Plug in the values: $x = \frac{9.00\,\text{N}}{2.00\,\text{N/cm}}$ .
Calculate $x$ : Plug in the values: $x = \frac{9.00\,\text{N}}{2.00\,\text{N/cm}}$ .Calculate $x$ : $x = \frac{9.00\,\text{N}}{2.00\,\text{N/cm}} = 4.5\,\text{cm}$ .