Govt. Exams
Entrance Exams
Net force = 60 - 20 = 40 N. Acceleration = 40/10 = 4 m/s². Velocity = 0 + 4 × 5 = 20 m/s
Using v² = u² - 2as, 0 = 225 - 2a(50), therefore a = 225/100 = 2.25 m/s²
Total acceleration = 10/5 = 2 m/s². For 3 kg block: F = 3 × 2 = 6 N (contact force)
T - mg = ma, 500 = m(10 + 2) = 12m, therefore m = 500/12 ≈ 41.67 kg, closest is 40 kg
Total mass = 75 kg. Acceleration = 10/5 = 2 m/s². Force = 75 × 2 = 150 N
On a frictionless incline, a = g sin θ = 10 × sin 30° = 10 × 0.5 = 5 m/s²
Friction f = μmg = 0.2 × 5 × 10 = 10 N. Net force = ma = 5 × 2 = 10 N. Applied force = 10 + 10 = 20 N
F = ma, so a = F/m. If m₁:m₂ = 2:3 and F is same, then a₁:a₂ = 3:2 (inverse proportion)
In elastic collision between identical objects where one is at rest, they exchange velocities. The moving ball comes to rest and the stationary ball moves with initial velocity of the first.
Max acceleration occurs at maximum extension. F = kx = 100×0.2 = 20 N. Using F = ma: 20 = m×50, m = 0.4 kg
