Example 1: A 5 kg block A rests on a frictionless table and is connected by a light string over a pulley to a 3 kg block B hanging vertically. Find the acceleration of the system and the tension in the string. Take g = 9.8 m/s2. Solution: 1. Free body diagram + Law 1/3: For block A, normal force balances its weight. Tension T pulls it right. For block B, weight mB g pulls down, tension T pulls up. Action-reaction: the string has the same T on both sides. 2. Apply F = ma to each block: - Block A: T = mA a - Block B: mB g - T = mB a 3. Solve: Add both equations → mB g = (mA + mB)a a = (mB g)/(mA + mB) = (3 x 9.8)/(5 + 3) = (29.4)/(8) = 3.68 m/s2 T = mA a = 5 x 3.68 = 18.4 N Insight: The 3kg block doesn’t fall at 9.8 m/s2 because the 5kg block resists it via tension. Example 2: A 10 kg box slides down a 30° frictionless incline. Find its acceleration. Solution: 1. Resolve forces: Gravity mg acts vertically down. Resolve into components parallel and perpendicular to the plane: - Parallel: mg 30^ - Perpendicular: mg 30^ — balanced by normal force, so no motion here. 2. Apply F = ma along the plane using Law 2: mg 30^ = m a a = g 30^ = 9.8 x 0.5 = 4.9 m/s2 Insight: Mass cancels out — all objects slide down a frictionless slope with the same acceleration. That’s Law 1: without friction, nothing else stops the motion. Example 3: A rocket ejects gas at 2000 m/s relative to the rocket at a rate of 50 kg/s. What is the thrust force on the rocket? Solution: 1. Law 3: The force on the gas = force on the rocket. This is called thrust. 2. Thrust = rate of change of momentum of ejected gas: F = ( p)/( t) = m v = 50 kg/s x 2000 m/s = 100,000 N Insight: No air is needed. The rocket pushes gas one way, gas pushes rocket the other way. Exercise a. A 2kg block is pushed across a rough floor with a 20N force. Friction is 8N. What is the acceleration Hint: Net force = Applied force - Friction = 20 - 8 = 12N. Then a = F/m = 12/2 = 6 m/s2._

Here's the solution to your exercise: a) A 2kg block is pushed across a rough floor with a 20N force. Friction is 8N. What is the acceleration? Step 1: Identify the forces acting on the block in the direction of motion. The applied force is 20 \, N. The friction force opposing the motion is 8 \, N. Step 2: Calculate the net force (F_net) acting on the block. F_net = F_applied - F_friction F_net = 20 \, N - 8 \, N F_net = 12 \, N Step 3: Apply Newton's Second Law (F_net = ma) to find the acceleration (a). Given mass m = 2 \, kg. a = F_netm a = 12 \, N2 \, kg a = 6 \, m/s^2 The acceleration of the block is 6 m/s^2. That's 2 down. 3 left today — send the next one.