the breaking strength of a steel cable is 20kN. If one pulls horizontally with this cable, what is the maximum horizontal acceleration which can be given to an 8 ton body resting on a rough horizontal surface if the coefficient of kinetic friction is 0.15

Options

A. `1.03 ms^-2`
B. `4.02 ms^-2`
C. `2.98 ms^-2`
D. `3.90 ms^-2`

Explanation

To solve the problem, we need to find the maximum horizontal acceleration that can be given to an 8-ton body resting on a rough horizontal surface, given the breaking strength of a steel cable and the coefficient of kinetic friction.

### Step 1: Convert the mass of the body from tons to kilograms.
1 ton = 1000 kg, so:
\[ \text{Mass (M)} = 8 \text{ tons} = 8 \times 1000 \text{ kg} = 8000 \text{ kg} \]

### Step 2: Identify the maximum pulling force (Tmax).
The breaking strength of the steel cable is given as 20 kN:
\[ T_{\text{max}} = 20 \text{ kN} = 20,000 \text{ N} \]

### Step 3: Calculate the normal force (N).
Since the body is resting on a horizontal surface, the normal force (N) is equal to the weight of the body:
\[ N = M \cdot g \]
Where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \):
\[ N = 8000 \text{ kg} \times 9.8 \text{ m/s}^2 = 78,400 \text{ N} \]

### Step 4: Calculate the kinetic frictional force (F_k).
The kinetic frictional force can be calculated using the formula:
\[ F_k = \mu_k \cdot N \]
Where \( \mu_k = 0.15 \):
\[ F_k = 0.15 \times 78,400 \text{ N} = 11,760 \text{ N} \]

### Step 5: Apply Newton's second law to find the maximum acceleration (A_max).
The net force acting on the body in the horizontal direction is given by:
\[ F_{\text{net}} = T_{\text{max}} - F_k \]
According to Newton's second law:
\[ F_{\text{net}} = M \cdot A_{\text{max}} \]
Thus, we can write:
\[ T_{\text{max}} - F_k = M \cdot A_{\text{max}} \]
Rearranging gives:
\[ A_{\text{max}} = \frac{T_{\text{max}} - F_k}{M} \]

### Step 6: Substitute the values to calculate A_max.
Substituting the known values:
\[ A_{\text{max}} = \frac{20,000 \text{ N} - 11,760 \text{ N}}{8000 \text{ kg}} \]
Calculating the numerator:
\[ 20,000 \text{ N} - 11,760 \text{ N} = 8,240 \text{ N} \]
Now, substituting this into the equation for acceleration:
\[ A_{\text{max}} = \frac{8,240 \text{ N}}{8000 \text{ kg}} = 1.03 \text{ m/s}^2 \]

### Final Answer:
The maximum horizontal acceleration that can be given to the body is:
\[ A_{\text{max}} = 1.03 \text{ m/s}^2 \]

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