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/s^2. 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/s^2 T = mA a = 5 x 3.68 = 18.4 N Insight: The 3kg block doesn’t fall at 9.8 m/s^2 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/s^2 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/s^2._

Question

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/s^2. 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/s^2 T = mA a = 5 x 3.68 = 18.4 N Insight: The 3kg block doesn’t fall at 9.8 m/s^2 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/s^2 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/s^2._

Asked on April 24, 2026Geography

This geography problem is solved step by step below, with detailed explanations to help you understand the method and arrive at the correct answer.

Accepted Answer · 2026-04-24T21:04:53.290Z

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 $20 \, \text{N}$. The friction force opposing the motion is $8 \, \text{N}$. Step 2: Calculate the net force ($F_{\text{net}}$) acting on the block. $$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$$ $$F_{\text{net}} = 20 \, \text{N} - 8 \, \text{N}$$ $$F_{\text{net}} = 12 \, \text{N}$$ Step 3: Apply Newton's Second Law ($F_{\text{net}} = ma$) to find the acceleration ($a$). Given mass $m = 2 \, \text{kg}$. $$a = \frac{F_{\text{net}}}{m}$$ $$a = \frac{12 \, \text{N}}{2 \, \text{kg}}$$ $$a = 6 \, \text{m/s}^2$$ The acceleration of the block is $\boxed{\text{6 m/s}^2}$. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-04-24T21:04:53.290Z

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 $20 \, \text{N}$. The friction force opposing the motion is $8 \, \text{N}$. Step 2: Calculate the net force ($F_{\text{net}}$) acting on the block. $$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$$ $$F_{\text{net}} = 20 \, \text{N} - 8 \, \text{N}$$ $$F_{\text{net}} = 12 \, \text{N}$$ Step 3: Apply Newton's Second Law ($F_{\text{net}} = ma$) to find the acceleration ($a$). Given mass $m = 2 \, \text{kg}$. $$a = \frac{F_{\text{net}}}{m}$$ $$a = \frac{12 \, \text{N}}{2 \, \text{kg}}$$ $$a = 6 \, \text{m/s}^2$$ The acceleration of the block is $\boxed{\text{6 m/s}^2}$. That's 2 down. 3 left today — send the next one.

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