Govt. Exams
Entrance Exams
Using v² = u² + 2as, with same deceleration, the object with higher initial velocity (A) travels a greater distance.
Tension T = m(g + a). Maximum T = 60 N. So 60 = 5(10 + a), which gives 12 = 10 + a, therefore a = 2 m/s².
Normal force N = mg = 10 × 10 = 100 N. Friction force f = μₖN = 0.2 × 100 = 20 N. Net force = 100 - 20 = 80 N. Acceleration = 80/10 = 8 m/s².
In uniform circular motion, acceleration (centripetal) is always directed toward the center of the circle (radially inward).
Acceleration a = (v-u)/t = (0-5)/0.1 = -50 m/s². Force F = ma = 2 × (-50) = -100 N. Magnitude = 100 N.
Apparent weight = m(g - a) = 50(10 - 2) = 50 × 8 = 400 N.
Acceleration a = (20-0)/10 = 2 m/s². Net force = ma = 1000 × 2 = 2000 N. Driving force = Net force + friction = 2000 + 500 = 2500 N.
Total acceleration a = F/(m₁+m₂) = 10/5 = 2 m/s². For block m₂, tension T = m₂ × a = 2 × 2 = 4 N.
The component of gravitational force along the incline is mg sin θ. Using F = ma, we get a = g sin θ.