NEET 2025 Q2: Stopping Distance Problems: Applying Work-Energy Theorem — Full solution + Why they asked this

PYQ + trends

1. TITLE

NEET 2025 Q2: Stopping Distance Problems: Applying Work-Energy Theorem — Full solution + Why they asked this

2. THE ACTUAL NEET QUESTION

NEET 2025 — Question
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m . If $F_A$ and $F_B$ are the forces applied by the breaks on cars A and B respectively, then the ratio of $\frac{F_A}{F_B}$ is

A) $\frac{1}{3}$
B) $\frac{1}{2}$
C) $\frac{3}{2}$
D) $\frac{2}{3}$

3. OFFICIAL ANSWER

NEET Answer Key: (D)

4. COMPLETE SOLUTION

• Step-by-step calculation:
  According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy ($W_{net} = \Delta KE$).
  Since the braking force $F$ acts in the opposite direction to the displacement $S$, the work done is $-FS$.
  The cars come to a complete stop, so final kinetic energy ($K_f$) is zero.
  Therefore: $-FS = 0 - K_i \Rightarrow FS = K_i$.

For car A: $F_A \times S_A = K_A \Rightarrow F_A = \frac{K_A}{S_A}$
For car B: $F_B \times S_B = K_B \Rightarrow F_B = \frac{K_B}{S_B}$

Now, take the ratio:
$\frac{F_A}{F_B} = \frac{K_A}{K_B} \times \frac{S_B}{S_A}$

Substitute the given values ($K_A = 100$, $K_B = 225$, $S_A = 1000$, $S_B = 1500$):
$\frac{F_A}{F_B} = \frac{100}{225} \times \frac{1500}{1000}$
$\frac{F_A}{F_B} = \frac{150}{225} = \frac{2}{3}$

• Formula used:
  Work-Energy Theorem: $W = \Delta K$ ($K_f - K_i = W$) and the scalar product for work $W = \vec{F} \cdot \vec{S} = FS \cos \theta$.

• Why this approach:
  It creates a direct mathematical link between braking force, stopping distance, and initial kinetic energy. Using the Work-Energy theorem provides a one-step scalar shortcut, completely bypassing the need for complex kinematics or calculating the deceleration/time.

5. WHY NEET ASKED THIS

• What concept were they testing? The examiner was testing the application of the Work-Energy Theorem for negative work scenarios (where $\theta=180^{\circ}$ and force opposes displacement).
• What's the common mistake they expected students to make? Students often try to forcefully calculate mass ($m$), initial velocity ($v$), and acceleration ($a$) using equations of motion ($v^2 = u^2 + 2as$ and $K = \frac{1}{2}mv^2$). Since exact masses aren't provided (only "similar cars"), students get stuck or waste precious exam time calculating variables that eventually cancel out.
• Which option was the "trap"? Option (C) $\frac{3}{2}$ is the classic reciprocal trap. If a student mistakenly assumes force is directly proportional to stopping distance ($F \propto S$) instead of inversely proportional ($F \propto \frac{1}{S}$), they will flip the final fraction.

6. TREND ANALYSIS

Prediction for NEET 2026:
Expect a [Stopping Distance / Work-Energy] question combined with variable resistive forces (e.g., stopping a block where friction increases linearly with distance) or asking to compare the stopping distances of two vehicles moving with the same momentum instead of the same kinetic energy.

7. SIMILAR QUESTIONS TO PRACTICE

"If you understood this, you can solve:"
- NEET 2024 Re Q3: Finding final kinetic energy after displacing an object with a vector force $\vec{F}=(-2\hat{i}+3\hat{j})$.
- NEET 2017 Q21: Using the Work-Energy theorem to find work done by air resistance on a falling raindrop.
- Stopping Distance with coefficient of friction ($\mu$): Calculating stopping distance when the braking force is replaced entirely by kinetic friction ($F_r = \mu mg$).

8. QUICK REVISION FOR THIS TYPE

For Stopping Distance Problems: Applying Work-Energy Theorem questions, always:

• [First check] if the final velocity is zero (meaning the object stops completely, making $K_f = 0$).
• [Then apply] the Work-Energy Theorem directly: Initial Kinetic Energy = Magnitude of Work Done by Braking Force ($K_i = F \times S$).
• [Watch out for] signs! The braking force always does negative work because it acts opposite to the displacement ($\theta = 180^\circ$, so $W = -FS$). This negative work perfectly cancels the negative change in kinetic energy ($0 - K_i$).

9. NCERT BASE

This question tests: NCERT Class 11, Chapter 5 (Work, Energy and Power), Page 74 — Specifically drawing from Example 5.3 (A cyclist comes to a skidding stop in 10 m under a 200 N opposing force) which explicitly demonstrates finding negative work done by a stopping force using the WE theorem.

10. CTA

All PYQs from this chapter → Work, Energy and Power Guide
Practice similar questions → [PYQ Test: https://knowmyexam.in/mock-tests]