Question

a horizontal force of 80 n is required to slide a 17 kg box across a tile floor at constant velocity. what is the coefficient of sliding friction between the box and the floor? 0.78 0.21 0.48 0.67

Explanation:

Step1: Recall the formula for sliding friction

The formula for sliding (kinetic) friction is \( F_f = \mu_k F_N \), where \( F_f \) is the frictional force, \( \mu_k \) is the coefficient of kinetic friction, and \( F_N \) is the normal force. When an object is on a horizontal surface and moving at constant velocity (so net force is zero), the applied horizontal force \( F_{app} \) equals the frictional force \( F_f \), and the normal force \( F_N \) equals the weight of the object \( mg \) (where \( m \) is mass and \( g \) is acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)).

Step2: Calculate the normal force

The mass of the box \( m = 17 \, \text{kg} \), so the weight (and thus the normal force) is \( F_N = mg = 17 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 166.6 \, \text{N} \).

Step3: Relate applied force to frictional force and solve for \( \mu_k \)

Since the box moves at constant velocity, the applied force \( F_{app} = 80 \, \text{N} \) equals the frictional force \( F_f \). From \( F_f = \mu_k F_N \), we can solve for \( \mu_k \):
\( \mu_k = \frac{F_f}{F_N} = \frac{F_{app}}{F_N} \) (because \( F_f = F_{app} \) for constant velocity).

Substituting the values: \( \mu_k = \frac{80 \, \text{N}}{166.6 \, \text{N}} \approx 0.48 \).

Answer:

0.48