Question

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying brakes, 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 brakes on cars A and B, respectively, then the ratio \( \frac{F_A}{F_B} \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{3} \)

Hint:

Remember the work-energy theorem: the work done on an object is equal to the change in its kinetic energy. Here, the final kinetic energy is zero, so the work done by the braking force is equal to the negative of the initial kinetic energy, or the magnitude of the work done is equal to the initial kinetic energy.

Answer 1

The Correct Option is C

Step 1: Work Done by the Braking Force
The work done by a constant force \(F\) over a distance \(d\) is given by:
\( W = Fd \).
The work done by the braking force to stop a car is equal to the initial kinetic energy of the car.

For Car A:
Initial kinetic energy \( KE_A = 100 \) J.
Stopping distance \( d_A = 1000 \) m.
Work done by braking force \( W_A = F_A d_A = F_A \times 1000 \) J.
Since work done equals initial kinetic energy, we have:
\( F_A \times 1000 = 100 \implies F_A = \frac{100}{1000} = 0.1 \) N.

For Car B:
Initial kinetic energy \( KE_B = 225 \) J.
Stopping distance \( d_B = 1500 \) m.
Work done by braking force \( W_B = F_B d_B = F_B \times 1500 \) J.
Since work done equals initial kinetic energy, we have:
\( F_B \times 1500 = 225 \implies F_B = \frac{225}{1500} \) N.

Step 2: Find the Ratio \( \frac{F_A}{F_B} \)
Now, let's find the ratio \( \frac{F_A}{F_B} \):
\[ \frac{F_A}{F_B} = \frac{0.1}{\frac{225}{1500}} = \frac{0.1 \times 1500}{225} = \frac{150}{225} \] Simplify the fraction:
\[ \frac{150}{225} = \frac{50 \times 3}{75 \times 3} = \frac{50}{75} = \frac{25 \times 2}{25 \times 3} = \frac{2}{3} \] Therefore, the ratio \( \frac{F_A}{F_B} = \frac{2}{3} \).