Physics Q5 – NEET-UG 2025

Home Physics Work, Energy & PowerQuestion #650

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

F_{A} / F_{B}

Clue 1 of 3: Work-Energy Theorem

The work done by a force is equal to the change in kinetic energy of the object.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Relate Work Done to Kinetic Energy✦ Active

According to the Work-Energy Theorem, the work done by the net force on an object is equal to the change in its kinetic energy. When a car stops, the final kinetic energy is zero. Therefore, the work done by the braking force is equal to the initial kinetic energy of the car.

W = Δ K = K_{i n i t i a l} - K_{f i n a l}

W = F \times d

F \times d = K_{i n i t i a l}

Step 2: Express Braking Forces for Each Car○ Expand

Using the relationship $F \times d = K$ , we can express the braking force for each car:

F_{A} = \frac{K_{A}}{d_{A}}

F_{B} = \frac{K_{B}}{d_{B}}

Given values are: $K_{A} = 100 J$ , $d_{A} = 1000 m$ , $K_{B} = 225 J$ , $d_{B} = 1500 m$ .

💡 Teacher's Secret Hint

Ensure units are consistent, though in this ratio problem, they will cancel out.

Step 3: Calculate the Ratio

F_{A} / F_{B}

○ Expand

Now, we calculate the ratio of the forces $F_{A} / F_{B}$ :

\frac{F_{A}}{F_{B}} = \frac{K_{A} / d_{A}}{K_{B} / d_{B}} = \frac{K_{A}}{d_{A}} \times \frac{d_{B}}{K_{B}}

= \frac{100 J}{1000 m} \times \frac{1500 m}{225 J}

= \frac{1}{10} \times \frac{1500}{225}

= \frac{150}{225}

= \frac{2 \times 75}{3 \times 75} = \frac{2}{3}

💡 Teacher's Secret Hint

Simplify the fraction by finding common factors. Both 150 and 225 are divisible by 25, and then by 3.

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Physics Question 5 – NEET-UG 2025

Clue 1 of 3: Work-Energy Theorem

🧠 Full Solution Path