A car of mass 1000 kg is moving with a speed of `40 ms^(-1)` on a circular path of radius 400m. If its speed is increasing at the rate of `3ms^(-2)` the total force acting on the car is

Options

A. 3000N
B. 4000 N
C. 5000 N
D. 7000 N

Answer

To find the total force acting on the car, we need to consider both the tangential and centripetal accelerations. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Mass of the car, \( m = 1000 \, \text{kg} \)
- Speed of the car, \( v = 40 \, \text{m/s} \)
- Radius of the circular path, \( r = 400 \, \text{m} \)
- Tangential acceleration, \( a_t = 3 \, \text{m/s}^2 \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(40 \, \text{m/s})^2}{400 \, \text{m}} = \frac{1600}{400} = 4 \, \text{m/s}^2 \]

### Step 3: Calculate the resultant acceleration
The total acceleration (\( a \)) is the vector sum of the tangential acceleration and the centripetal acceleration. We can use the Pythagorean theorem since these two accelerations are perpendicular to each other:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values:
\[ a = \sqrt{(3 \, \text{m/s}^2)^2 + (4 \, \text{m/s}^2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Step 4: Calculate the total force acting on the car
Using Newton's second law, the total force (\( F \)) can be calculated as:
\[ F = m \cdot a \]
Substituting the values:
\[ F = 1000 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 5000 \, \text{N} \]

### Final Answer
The total force acting on the car is \( 5000 \, \text{N} \).

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