--- title: "How hard do you need to push?" subtitle: "Year 10 Physics · Newton's laws · Lesson 4" --- # How hard do you need to push? **Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}} > **Key question:** Push an empty shopping trolley and it shoots off. Push a full one just as hard and it barely gets going. Same push, different result: by exactly how much? :::teacher-note **In one sentence:** Students apply F = ma to find force, mass or acceleration with units, use the net force, and explain how changing force or mass changes acceleration (AC9S10U05, node 4, a hinge node). **Students should be able to say:** I can use F = ma to find force, mass or acceleration, with correct units. I can explain that F in F = ma is the net force. I can explain what doubling the force or the mass does to the acceleration. **Width:** the unmarked parts are the floor every student does (Engage, Explore, Explain, Evaluate). Typical and Wider sections add contexts, not difficulty. Students work in the online tutorial alongside this sheet. ::: ## Engage :::keep-together **A push gives an empty trolley an acceleration of 4 m/s². You load it until its mass is doubled and push exactly as hard. Predict its acceleration.** ☐ 4 m/s²: the push is the same ☐ 8 m/s² ☐ A little less, maybe 3.5 m/s² ☐ 2 m/s²: half as much My prediction, and why: {{lines-3}} ::: :::teacher-note **What happens:** 2 m/s². Doubling the mass halves the acceleration for the same net force: a = F ÷ m. ::: ## Explore Try it on the board in the tutorial. Was your prediction right? What was different? {{lines-2}} ## Explain: F = ma :::keep-together ### The rule **For 4 kg you needed 8 N; for 8 kg, 16 N. What rule links force, mass and acceleration?** ☐ Force = mass + acceleration ☐ Force = mass ÷ acceleration ☐ Force = mass × acceleration (F = ma) ::: :::teacher-note **Answer:** Force = mass × acceleration (F = ma) 8 = 4 × 2 and 16 = 8 × 2. F is the net force in N, m the mass in kg, a the acceleration in m/s². ::: :::keep-together ### Find the force **What net force gives a 1200 kg car an acceleration of 3 m/s², in N?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 3600 F = ma = 1200 × 3 = 3600 N. ::: :::keep-together ### Find the acceleration **A net force of 50 N acts on a 10 kg box. Its acceleration, in m/s²?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 5 a = F ÷ m = 50 ÷ 10 = 5 m/s². ::: :::keep-together ### Find the mass **A net force of 600 N gives a sprinter an acceleration of 10 m/s². Her mass, in kg?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 60 m = F ÷ a = 600 ÷ 10 = 60 kg. ::: :::keep-together ### Net force **A 2 kg box is pushed with 10 N against 4 N of friction. What force goes into F = ma?** ☐ The friction, 4 N ☐ The net force, 6 N ☐ The push, 10 N ☐ 14 N ::: :::teacher-note **Answer:** The net force, 6 N Net force 10 − 4 = 6 N, so a = 6 ÷ 2 = 3 m/s². ::: :::keep-together ### Units **Match each quantity to its unit.** A. newtons (N) B. metres per second (m/s) C. kilograms (kg) D. metres per second squared (m/s²) {{columns 60% 40%}} {{rows 1}} | Part | Letter | |---|---| | `Force (F)` | | | `Mass (m)` | | | `Acceleration (a)` | | | `Velocity (v)` | | ::: :::teacher-note **Answer:** Force (F) → A. newtons (N); Mass (m) → C. kilograms (kg); Acceleration (a) → D. metres per second squared (m/s²); Velocity (v) → B. metres per second (m/s) F = ma only works with these units: mass in kg (not g), acceleration in m/s² (not m/s). ::: ## proportions :::keep-together ### Double, half or same **What happens to the acceleration?** Groups: Doubles, Halves, Stays the same - The net force doubles, same mass: {{field-15}} - The mass doubles, same net force: {{field-15}} - Both force and mass double: {{field-15}} - The net force halves, same mass: {{field-15}} ::: :::teacher-note **Answer:** The net force doubles, same mass → Doubles; The mass doubles, same net force → Halves; Both force and mass double → Stays the same; The net force halves, same mass → Halves ::: :::keep-together ### Triple the force **A trolley accelerates at 6 m/s². The net force is tripled and the mass stays the same. New acceleration, in m/s²?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 18 Three times the force, three times the acceleration: 18 m/s². ::: :::typical ## Elaborate: stopping a car :::keep-together ### Deceleration **A 1500 kg car brakes from 20 m/s to rest in 4 s. What is the size of its deceleration, in m/s²?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 5 Δv ÷ Δt = 20 ÷ 4 = 5 m/s² (backwards). ::: :::keep-together ### Braking force **What braking force does that need, in N?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 7500 F = ma = 1500 × 5 = 7500 N, backwards. ::: :::keep-together ### A loaded car **The same car is loaded with people and luggage (more mass). To stop in the same time, the braking force must be…** ☐ the same ☐ smaller ☐ bigger ::: :::teacher-note **Answer:** bigger F = ma: the same deceleration with more mass needs more force. With the same braking force, the loaded car would decelerate less and take longer to stop. ::: ::: :::wider ## Elaborate: is it really a straight line? :::keep-together **You measure a trolley's acceleration for several pulling forces, keeping the mass the same. What graph of acceleration against force does F = ma predict?** ☐ A curve that levels off ☐ A straight line through the origin (gradient 1 ÷ mass) ☐ A horizontal line ::: :::teacher-note **Answer:** A straight line through the origin (gradient 1 ÷ mass) a = F ÷ m, so a is proportional to F: a straight line through the origin with gradient 1/m. Real data that misses the origin usually points to friction. ::: ::: ## Evaluate: look back Write one sentence: "At first I thought… Now I think… because…" **Revision summary** - **Newton's second law:** **F = ma**. F is the **net** force (N), m the mass (kg), a the acceleration (m/s²). - Rearranged: a = F ÷ m and m = F ÷ a. - Double the net force, double the acceleration. Double the mass, half the acceleration. - The acceleration points the same way as the net force. ## Exit ticket :::keep-together **What acceleration does a net force of 30 N give a 6 kg trolley?** {{lines-3}} ::: :::teacher-note **Answer:** a = F ÷ m = 30 ÷ 6 = 5 m/s², in the direction of the net force. :::