Question

section 1
read the passage and answer the following question(s).
passage 9026
directions: the graphic shows a box at rest on a rough horizontal surface. use the graphic to answer any questions that follow.

3 once the box is moving, it requires 140 newtons (n) to maintain a constant acceleration. what is the coefficient of kinetic friction, , between the box and the ground?
a 0.43
b 0.24
c 0.34
d 0.30

Explanation:

Step1: Recall the formula for kinetic friction

The force of kinetic friction \( F_f \) is given by \( F_f = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. On a horizontal surface, the normal force \( N \) is equal to the weight of the object, so \( N = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)).

Step2: Calculate the normal force

Given \( m = 47 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \), we calculate \( N \):

$$ N = mg = 47 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 460.6 \, \text{N} $$

Step3: Relate the applied force and friction force (constant acceleration)

When the box is moving with constant acceleration, the net force is zero (since acceleration is constant, \( F_{\text{net}} = ma = 0 \) implies \( F = F_f \)). So the force of kinetic friction \( F_f \) is equal to the applied force \( F = 140 \, \text{N} \).

Step4: Solve for the coefficient of kinetic friction

Using \( F_f = \mu_k N \), we can solve for \( \mu_k \):

$$ \mu_k = \frac{F_f}{N} = \frac{140 \, \text{N}}{460.6 \, \text{N}} \approx 0.30 $$

Answer:

D. 0.30