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. $10^4$ N
• B. $2 \times 10^4$ N
• C. $3 \times 10^4$ N
• D. $5 \times 10^3$ N
• E. $10^3$ N

Hint:

For problems involving force, distance, and changes in speed (without mentioning time), the work-energy principle is often faster than using kinematic equations and Newton's laws separately . Also, always check that the velocity is in standard S.I. units (m/s) before squaring it in energy calculations. The applied average force opposes motion.

Answer 1

The Correct Option is B

Concept: The average force required to stop a vehicle can be determined using the Work-Energy Theorem, which states that the work done by the braking force is equal to the change in kinetic energy.
• Kinetic Energy: $K.E = \frac{1}{2}mv^2$
• Work Done: $W = F \times s$
• Work-Energy Principle: $F \times s = \Delta K.E$

Step 1: {Convert the initial velocity from km/hr to m/s.}
$$u = 36 \text{ km/hr} = 36 \times \frac{5}{18} \text{ m/s}$$ $$u = 2 \times 5 = 10 \text{ m/s}$$

Step 2: {List the mass and displacement values.}

• Mass ($m$) = 2000 kg
• Displacement ($s$) = 5 m
• Final velocity ($v$) = 0 m/s (stopped)

Step 3: {Set up the Work-Energy equation.}
The work done by the force $F$ over distance $s$ equals the initial kinetic energy: $$F \times s = \frac{1}{2}mu^2$$

Step 4: {Substitute the values and solve for Force.}
$$F \times 5 = \frac{1}{2} \times 2000 \times (10)^2$$ $$5F = 1000 \times 100$$ $$5F = 100000$$

Step 5: {Calculate the final average force.}
$$F = \frac{100000}{5}$$ $$F = 20000 \text{ N}$$ $$F = 2 \times 10^4 \text{ N}$$