Question:medium
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:
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:
Show 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.
Updated On: Mar 27, 2026
- \( \frac{1}{2} \)
- \( \frac{1}{4} \)
- \( \frac{2}{3} \)
- \( \frac{4}{3} \)
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Work Done by Braking Force
Work done by a constant force \(F\) over distance \(d\) is \( W = Fd \).
The work performed by the braking force to stop a car equals its initial kinetic energy.
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.
Work done equals initial kinetic energy:
\( F_A \times 1000 = 100 \implies F_A = \frac{100}{1000} = 0.1 \) N.
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.
Work done equals initial kinetic energy:
\( F_B \times 1500 = 225 \implies F_B = \frac{225}{1500} \) N.
Step 2: Calculate Ratio \( \frac{F_A}{F_B} \)
Calculate 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:
\[ \frac{150}{225} = \frac{50 \times 3}{75 \times 3} = \frac{50}{75} = \frac{25 \times 2}{25 \times 3} = \frac{2}{3} \] The ratio \( \frac{F_A}{F_B} = \frac{2}{3} \).
Work done by a constant force \(F\) over distance \(d\) is \( W = Fd \).
The work performed by the braking force to stop a car equals its initial kinetic energy.
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.
Work done equals initial kinetic energy:
\( F_A \times 1000 = 100 \implies F_A = \frac{100}{1000} = 0.1 \) N.
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.
Work done equals initial kinetic energy:
\( F_B \times 1500 = 225 \implies F_B = \frac{225}{1500} \) N.
Step 2: Calculate Ratio \( \frac{F_A}{F_B} \)
Calculate 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:
\[ \frac{150}{225} = \frac{50 \times 3}{75 \times 3} = \frac{50}{75} = \frac{25 \times 2}{25 \times 3} = \frac{2}{3} \] The ratio \( \frac{F_A}{F_B} = \frac{2}{3} \).
Was this answer helpful?
7
