A force `F_(1)` of 500 N is required to push a car of mass 1000 kg slowly at constant speed on a leveled road. If a force `F_(2)` of 1000N is applied, the acceleration of the car will be `:`

Options

A. zero
B. `1.5ms^(-2)`
C. `1 ms^(-2)`
D. `0.5 ms^(-2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the forces acting on the car
- We have two forces acting on the car:
- The applied force \( F_2 = 1000 \, \text{N} \)
- The frictional force \( F_1 = 500 \, \text{N} \) (which is equal to the force required to move the car at a constant speed)

### Step 2: Calculate the net force acting on the car
- The net force \( F_{\text{net}} \) acting on the car can be calculated using the formula:
\[ F_{\text{net}} = F_2 - F_1 \]
- Substituting the values:
\[ F_{\text{net}} = 1000 \, \text{N} - 500 \, \text{N} = 500 \, \text{N} \]

### Step 3: Apply Newton's second law of motion
- According to Newton's second law, the net force is also equal to the product of mass and acceleration:
\[ F_{\text{net}} = m \cdot a \]
- Here, \( m = 1000 \, \text{kg} \) and \( a \) is the acceleration we need to find.

### Step 4: Rearranging the equation to find acceleration
- Rearranging the equation gives:
\[ a = \frac{F_{\text{net}}}{m} \]
- Substituting the values we found:
\[ a = \frac{500 \, \text{N}}{1000 \, \text{kg}} = 0.5 \, \text{m/s}^2 \]

### Step 5: Conclusion
- The acceleration of the car when a force of \( F_2 = 1000 \, \text{N} \) is applied is:
\[ a = 0.5 \, \text{m/s}^2 \]

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