Skip to content
PhysicsNEET 2025Work by constant and variable forces

Work, Energy and Power

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 FAF_A and FBF_B are the forces applied by the brakes on cars A and B respectively, then the ratio FAFB\dfrac{F_A}{F_B} is:

  1. A.32\dfrac{3}{2}
  2. B.23\dfrac{2}{3}✓
  3. C.13\dfrac{1}{3}
  4. D.12\dfrac{1}{2}

Solution

Work–energy: Fs=KEF s = KE. FAFB=KEA/sAKEB/sB=100/1000225/1500=0.10.15=23\frac{F_A}{F_B} = \frac{KE_A / s_A}{KE_B / s_B} = \frac{100/1000}{225/1500} = \frac{0.1}{0.15} = \frac{2}{3}.

Save for spaced revision (free account)

More from this chapter in past papers

From Physics › Work, Energy and Power › Work by constant and variable forces — lectures, notes and every past question on it.

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram