A vehicle of 100 kg is moving with a velocity of `5m//sec` . To stop it in `(1)/(10)`sec , the required force in opposite direction is

Options

A. 5000 N
B. 500 N
C. 50 N
D. 1000 N

Answer

To solve the problem of stopping a vehicle of mass 100 kg moving at a velocity of 5 m/s in \( \frac{1}{10} \) seconds, we can follow these steps:

### Step 1: Identify the initial conditions
- Mass of the vehicle, \( m = 100 \, \text{kg} \)
- Initial velocity, \( u = 5 \, \text{m/s} \)
- Final velocity, \( v = 0 \, \text{m/s} \) (since we want to stop the vehicle)
- Time to stop, \( t = \frac{1}{10} \, \text{s} = 0.1 \, \text{s} \)

### Step 2: Use the equation of motion
We can use the equation of motion:
\[ v = u + at \]
where \( a \) is the acceleration (or deceleration in this case).

### Step 3: Rearrange the equation to find acceleration
Substituting the known values into the equation:
\[ 0 = 5 + a \cdot 0.1 \]
Rearranging gives:
\[ a \cdot 0.1 = -5 \]
\[ a = \frac{-5}{0.1} = -50 \, \text{m/s}^2 \]

### Step 4: Calculate the required force
According to Newton's second law of motion, the force \( F \) required to produce this acceleration is given by:
\[ F = ma \]
Substituting the values:
\[ F = 100 \, \text{kg} \cdot (-50 \, \text{m/s}^2) \]
\[ F = -5000 \, \text{N} \]

### Step 5: Interpret the result
The negative sign indicates that the force is applied in the opposite direction to the motion of the vehicle. Therefore, the required force to stop the vehicle is:
\[ F = 5000 \, \text{N} \text{ (in the opposite direction)} \]

### Final Answer
The required force in the opposite direction is **5000 N**.

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