A 25.6 kg box moving on flat ground slows down and comes to a complete stop due to a frictional force.

The coefficient of kinetic friction is 0.124. What is the acceleration of the box?

[tex]\[ a = [?] \, \text{m/s}^2 \][/tex]



Answer :

To determine the acceleration of the box, we need to follow a series of steps involving the concepts of frictional force and Newton's second law of motion. Here is the detailed step-by-step solution:

### Step 1: Calculate the Normal Force
First, we calculate the normal force acting on the box. On flat ground, the normal force [tex]\( F_N \)[/tex] is equal to the gravitational force acting on the box. This can be calculated using the formula:
[tex]\[ F_N = m \cdot g \][/tex]
where:
- [tex]\( m = 25.6 \)[/tex] kg (mass of the box)
- [tex]\( g = 9.81 \)[/tex] m/s² (acceleration due to gravity)

### Step 2: Normal Force Calculation
[tex]\[ F_N = 25.6 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 251.136 \, \text{N} \][/tex]

### Step 3: Calculate the Frictional Force
Next, we need to calculate the kinetic frictional force [tex]\( F_f \)[/tex]. The frictional force is given by:
[tex]\[ F_f = \mu_k \cdot F_N \][/tex]
where:
- [tex]\( \mu_k = 0.124 \)[/tex] (coefficient of kinetic friction)

### Step 4: Frictional Force Calculation
[tex]\[ F_f = 0.124 \times 251.136 \, \text{N} = 31.140864 \, \text{N} \][/tex]

### Step 5: Determine the Acceleration
Finally, to determine the acceleration [tex]\( a \)[/tex] produced by the frictional force, we use Newton's second law:
[tex]\[ F = m \cdot a \][/tex]
Rearranging for acceleration [tex]\( a \)[/tex]:
[tex]\[ a = \frac{F}{m} \][/tex]
Since this is a deceleration due to friction, the force [tex]\( F \)[/tex] is the frictional force [tex]\( F_f \)[/tex], and it acts in the opposite direction of motion:

### Step 6: Acceleration Calculation
[tex]\[ a = -\frac{31.140864 \, \text{N}}{25.6 \, \text{kg}} = -1.21644 \, \text{m/s}^2 \][/tex]

### Conclusion
Thus, the acceleration of the box is:
[tex]\[ a = -1.21644 \, \text{m/s}^2 \][/tex]
The negative sign indicates that this is a deceleration.