Question

A vehicle moving at 36 km/hr is to be stopped by applying brakes in the next 5 m. If the vehicle weighs 2000 kg, determine the average force that must be applied on it.

• A. \( 10^4 \, \text{N} \)
• B. \( 2 \times 10^4 \, \text{N} \)
• C. \( 3 \times 10^4 \, \text{N} \)
• D. \( 5 \times 10^3 \, \text{N} \)
• E. \( 10^3 \, \text{N} \)

Hint:

To calculate the force required to stop a vehicle, use the equation of motion \( v^2 = u^2 + 2as \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the stopping distance. Then apply \( F = ma \) to find the force.

Answer 1

The Correct Option is B

Step 1: Given information.
- Initial velocity \( u = 36 \, \text{km/hr} = \frac{36 \times 1000}{3600} = 10 \, \text{m/s} \)
- Distance \( s = 5 \, \text{m} \)
- Mass of the vehicle \( m = 2000 \, \text{kg} \)

Step 2: Applying the equation of motion.
We can use the following equation of motion to find the acceleration \( a \): \[ v^2 = u^2 + 2as \] Since the vehicle is coming to rest, \( v = 0 \). Substituting the values: \[ 0 = u^2 + 2as \quad \Rightarrow \quad a = -\frac{u^2}{2s} \] Substituting \( u = 10 \, \text{m/s} \) and \( s = 5 \, \text{m} \): \[ a = -\frac{(10)^2}{2 \times 5} = -10 \, \text{m/s}^2 \]

Step 3: Finding the force.
The average force \( F \) can be calculated using Newton's second law: \[ F = ma \] Substituting \( m = 2000 \, \text{kg} \) and \( a = -10 \, \text{m/s}^2 \): \[ F = 2000 \times (-10) = -20000 \, \text{N} \] The magnitude of the force is \( 20000 \, \text{N} \), which corresponds to \( 2 \times 10^4 \, \text{N} \).

Final Answer: (B) \( 2 \times 10^4 \, \text{N} \)