Example 1: A 5 kg block A rests on a frictionless table and is connected by a light string over a pulley to a…

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/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 × 9.8)/(5 + 3) = 29.4/8 = 3.68 m/s² T = mA a = 5 × 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 × 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._

ScanSolve AI Tutor · Accepted Answer

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?

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 $8 \, \text{N}$ .
Calculate the net force ( $F_{\text{net}}$ ) acting on the block.
$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$ $F_{\text{net}} = 20 \, \text{N} - 8 \, \text{N}$ $F_{\text{net}} = 12 \, \text{N}$
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 $\boxed{\text{6 m/s}^2}$ .

Step-by-step solution

Identify the forces acting on the block in the direction of motion.

Calculate the net force ( $F_{\text{net}}$ ) acting on the block.

Apply Newton's Second Law ( $F_{\text{net}} = ma$ ) to find the acceleration ( $a$ ).

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