Govt. Exams
Entrance Exams
Net force = 5000 - 2000 = 3000 N. Using F = ma, a = 3000/1000 = 3 m/s²
At rest, normal force equals weight. N = mg, 10 = m × 10, therefore m = 1 kg
Net force = √(3² + 4²) = 5 N. Using F = ma, a = 5/2 = 2.5 m/s²
For constant velocity, acceleration = 0. By Newton's first law, net force = 0. Therefore, friction force must equal applied force = 50 N.
An inertial frame has zero acceleration. A rotating frame undergoes centripetal acceleration, making it non-inertial. All others have constant velocity (zero acceleration).
Using s = ut + ½at², where u = 0, s = 100 m, t = 10 s. Therefore, 100 = 0 + ½a(100), giving a = 2 m/s²
Using v = u + at: 20 = 10 + a(5), a = 2 m/s². F = ma = 2×2 = 4 N
Using v = u + at: 0 = 20 + a(5), a = -4 m/s². Deceleration = 4 m/s²
By conservation of energy and kinematics, the stone returns with the same speed but opposite direction: 30 m/s downward
Using s = ut + ½at², where u = 0, a = 5 m/s², t = 6s: s = 0 + ½(5)(36) = 90 m
