Question

question 3 of 6 policies current attempt in progress to measure the static friction coefficient between a 4.19 - kg block and a vertical wall, the setup shown in the drawing is used. a spring (spring constant = 458 n/m) is attached to the block. someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. if the spring in such a setup is compressed by 0.0537 m, what is the coefficient of static friction?

Explanation:

Step1: Calculate the normal force

According to Hooke's law \(F = kx\), where \(k = 458\space N/m\) and \(x=0.0537\space m\).
\(F_N=kx = 458\times0.0537\space N\)

Step2: Calculate the maximum static - friction force

The maximum static - friction force \(f_s=\mu_sF_N\). In the vertical direction, when the block does not slip, \(f_s = mg\), where \(m = 4.19\space kg\) and \(g = 9.8\space m/s^2\), so \(mg=\mu_sF_N\).
From \(F_N = kx\), we can express \(\mu_s=\frac{mg}{kx}\)
Substitute \(m = 4.19\space kg\), \(g = 9.8\space m/s^2\), \(k = 458\space N/m\) and \(x = 0.0537\space m\) into the formula:
\(\mu_s=\frac{4.19\times9.8}{458\times0.0537}\)
First, calculate the numerator: \(4.19\times9.8=41.062\)
Then, calculate the denominator: \(458\times0.0537 = 458\times(0.05 + 0.0037)=458\times0.05+458\times0.0037=22.9+1.6946 = 24.5946\)
\(\mu_s=\frac{41.062}{24.5946}\approx1.67\)

Answer:

\(\mu_s\approx1.67\)