Question

simple harmonic motion the relationship you developed is hooke’s law: $f_s = -kx$. given that the applied force is opposite the spring force, hooke’s law can be rewritten as $f_{\text{applied}} = kx$. if an applied force of 35 n causes a spring to undergo a displacement of 22 cm, what is its spring constant? enter your answer in the box. $k = \square$ n/m

Explanation:

Step1: Recall Hooke's Law formula

We know from the problem that the formula for the applied force in terms of spring constant \( k \) and displacement \( x \) is \( F_{\text{applied}} = kx \). We need to solve for \( k \), so we can rearrange the formula to \( k=\frac{F_{\text{applied}}}{x} \).

Step2: Convert units

The displacement \( x \) is given as 22 cm. Since the unit of the spring constant \( k \) is N/m, we need to convert centimeters to meters. We know that \( 1 \, \text{m} = 100 \, \text{cm} \), so \( x = 22 \, \text{cm}=\frac{22}{100} \, \text{m} = 0.22 \, \text{m} \).

Step3: Substitute values into the formula

We are given \( F_{\text{applied}} = 35 \, \text{N} \) and \( x = 0.22 \, \text{m} \). Substituting these values into the formula \( k=\frac{F_{\text{applied}}}{x} \), we get \( k = \frac{35}{0.22} \).

Step4: Calculate the value of \( k \)

Calculating \( \frac{35}{0.22}\approx159.09 \) (rounded to two decimal places).

Answer:

\( k \approx \boxed{159.09} \) N/m