Question

consider the system shown in the figure (figure 1). block a weighs 37.3 n and block b weighs 16.2 n. once block b is set into downward motion, it descends at a constant speed. calculate the coefficient of kinetic friction between block a and the tabletop.

Explanation:

Step1: Analyze forces on Block B (constant speed, so net force is zero)

Block B is moving at constant speed, so the tension \( T \) in the string equals its weight. So \( T = W_B = 16.2 \, \text{N} \).

Step2: Analyze forces on Block A (constant speed, so net force is zero)

For Block A, the tension \( T \) must equal the kinetic friction force \( f_k \). The kinetic friction force is given by \( f_k=\mu N \), where \( N \) is the normal force on Block A, which equals its weight \( W_A = 37.3 \, \text{N} \). So \( T=\mu W_A \).

Step3: Solve for \(\mu\)

Substitute \( T = 16.2 \, \text{N} \) and \( W_A = 37.3 \, \text{N} \) into \( T=\mu W_A \). We get \( \mu=\frac{T}{W_A}=\frac{16.2}{37.3} \).

Calculate \( \frac{16.2}{37.3}\approx0.434 \).

Answer:

\( \approx 0.434 \)