The breaking strength of the cable used to pull a body is 40 N. A body of mass 8 kg is resting on a table of coefficient of friction `mu=0.20`. The maximum acceleration which can be produced by the cable is ( take , `g=ms^(-2)` )

Options

A. `6 ms^(-2)`
B. `3 ms^(-2)`
C. `8 ms^(-2)`
D. `8 ms^(-2)`

Explanation

To find the maximum acceleration that can be produced by the cable, we can follow these steps:

### Step 1: Identify the forces acting on the body
The body is subjected to two main forces:
1. The tension (T) in the cable, which is 40 N (maximum breaking strength).
2. The frictional force (F_friction) opposing the motion.

### Step 2: Calculate the normal force (N)
The normal force is equal to the weight of the body since it is resting on a horizontal table. The weight (W) can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 8 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 8 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 78.4 \, \text{N} \]
Thus, the normal force \( N = 78.4 \, \text{N} \).

### Step 3: Calculate the frictional force (F_friction)
The frictional force can be calculated using the formula:
\[ F_friction = \mu \cdot N \]
Where:
- \( \mu = 0.20 \) (coefficient of friction)

Calculating the frictional force:
\[ F_friction = 0.20 \cdot 78.4 \, \text{N} = 15.68 \, \text{N} \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force (F_net) acting on the body is equal to the mass (m) multiplied by the acceleration (a):
\[ F_{net} = T - F_{friction} \]
Where:
- \( T = 40 \, \text{N} \) (tension in the cable)

Calculating the net force:
\[ F_{net} = 40 \, \text{N} - 15.68 \, \text{N} = 24.32 \, \text{N} \]

### Step 5: Calculate the acceleration (a)
Using Newton's second law:
\[ F_{net} = m \cdot a \]
Substituting the values:
\[ 24.32 \, \text{N} = 8 \, \text{kg} \cdot a \]

Solving for acceleration:
\[ a = \frac{24.32 \, \text{N}}{8 \, \text{kg}} = 3.04 \, \text{m/s}^2 \]

### Conclusion
The maximum acceleration that can be produced by the cable is approximately:
\[ a \approx 3.04 \, \text{m/s}^2 \]