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AS Mechanics Unit 2 (PHYA2) NI Newton’s Laws • Objects stay at rest or in uniform motion (velocity, i.e., speed + direction) unless acted on by an unbalanced force. AR constant v W Notes: • Add up the forces in each perpendicular direction (x, y, z) and see if there is a resultant – i.e., non-zero force Fx, Fy or Fz. • The Ancient Greeks believed bodies’ usual state was to be still. They didn’t understand friction. Newton & Galileo showed that moving bodies stay moving. Think satellites and airtracks… NII Newton’s Laws • The force on a body is proportional to its rate of change of momentum. Notes: • We have defined the newton (N) so that ∑F = ∆p/∆t • Since p = mv, ∑F = ∆mv/∆t we have two cases (a) m constant: ∑F = m∆v/∆t = ma (b) v constant: ∑F = v∆m/∆t • Examples: space rocket, tractor pulling trailer, water through hole in reservoir wall, winch. Case (a) or (b)? • This applies in each perpendicular direction (x, y, z), i.e., ∑Fx = max, ∑Fy = vy∆m/∆t, etc., etc. Newton’s Laws • “Objects stay at rest or move with uniform velocity unless acted on by a force” – No forces – Balanced forces (i.e., resultant = 0) NI • Bigger forces cause bigger accelerations, larger masses get smaller accelerations, hence F = ma – F is the RESULTANT force (F1 + F2 + ...) – a is ALWAYS in the same direction as F (even if v isn’t) • The rate of change of momentum of a body is NII proportional to the resultant force on it. – Hence F ~ mv/t – 1 N is defined so that it is ‘equal’ as well as ‘proportional’ NIII Newton’s Laws • When body A exerts a force on body B, then B exerts an equal (magnitude) and opposite (direction) a force on A. Notes: • FA–B = –FB–A • 6 pairs to find here: • Four conditions to be a NIII pair (a) same type of force, (b) same magnitude, (c) opposite directions, (d) acting on different objects. Weight • Weight is a force • F = ma, W = mg g: • the gravitational field strength, or • the acceleration due to gravity • is independent of m • is 9.81 Nkg–1 (or 9.81 ms–2) in London Classic mechanics problems • Lifting mass Classic mechanics problems • Man in a lift – Force on lift = T – mg = ma – If lift is stationary or travelling at constant velocity, T = mg Change in motion Direction of travel accelerating decelerating Up T>mg T<mg Down T<mg T<mg Classic mechanics problems • Masses over a frictionless pulley (Atwood’s machine) • Tension in rope T is same on both sides • m2g – T = m2a • T – m1g = m1a • (adding) m2g – m1g = m2a + m1a • or m2 m1 a g m2 m1 • Can think of this as both masses being accelerated by a force equal to the difference in their weights T T Classic mechanics problems • Inclined slope, no friction • Force acting down slope is mgsinq • So a = gsinq • Block slides down slope • Note that although it takes longer to reach the bottom than if it had dropped vertically, its speed at the bottom is the same – can you explain why? Classic mechanics problems • Inclined slope, with friction (F) • Force acting down slope is mgsinq – F • So mgsinq – F = ma • This time the velocity at the bottom of the slope will be less than the frictionless case. • Try the questions on p.137 Terminal speed • Speed or velocity as we’re only interested in the downwards direction • Drag depends on the object’s shape and speed, and the viscosity of the fluid • Maximum acceleration is when v = 0. Use the gradient of the v–t graph. Counter force Horizontal motion Driving force Driving force – provided by rider/engine Counter force – air resistance and friction • Driving force < counter force: vehicle slows down • Driving force = counter force: vehicle moves at constant velocity • Driving force > counter force: vehicle speeds up Stopping distances • stopping = thinking + braking s = s1 + s2 s1 = vt0 (speed × reaction time) s2 = v2/2a (from v2 = u2 + 2as) (or s2 = WD/Fbraking) • Thinking distance a v • Braking distance a v2 Impact time • Impact time is the duration of an impact force • Remember s u v t 2 Average speed 2s • so impact time is t uv • acceleration is vu a t Impact example • A 500 kg car crashes into a lamp post. The car stops from a speed of 13 ms–1 in a distance of 1 m. 2s 2 t 0.15s • t=? u v 13 v u 13 2 a 86 . 7 ms 8.8 g • a=? t 0.15 • F=? F ma 500 86.7 43350N Road safety • Airbags inflate to slow the deceleration of the head and spread pressure over a wider area • Seatbelts are designed to stretch, slowing the deceleration of the body • Crumple zones in vehicles increase the deceleration time • Padding inside crash helmet increases deceleration time • Now try Qs on p. 145 Work • When you expend energy to exert a force which moves an object, you are doing “work”. work done Force x distance in the direction of the force Joules Newtons metres • Work done = energy transferred Is any work being done in these cases? Work questions • How much work is done when a mass of 3 kg is lifted vertically through 6 m? – Work = F × d = (3 × 10) × 6 = 180 J • A hiker climbs a hill 300 m high. If she weighs 500 N calculate the work she does lifting her body to the top of the hill. – Work = F × d = 500 × 300 = 150,000 J or 150 kJ Motion and force in different directions • Work done = component of average force in direction of motion × distance moved in direction of force • W = Fscosq • If q = 90°, W = 0. Force–distance graphs • Area under a force–distance graph is work done Constant force varying force Force extending a spring F = kx • So work to extend to L is area under graph 1 W L F 2 force • The more you stretch it, the harder you have to pull • Hooke’s Law: F L extension Kinetic energy • KE is the result of work being done. • Imagine a constant force F accelerating a mass m: t=0, u=0 v F m m uv 1 v u v s t vt, a 2 2 t t mv F ma t mv vt 1 2 W Fs mv t 2 2 Kinetic Energy • Kinetic energy (KE) is the energy an object has due to its motion. • So an object has more KE if: 1 2 KE mv 2 mass – it has a greater mass, or – it moves faster • A lorry and a van both travel at 15 ms–1. The lorry has a mass of 2000 kg and the van a mass of 1000 kg. What KE does each have? 30 kJ 15 kJ velocity Kinetic Energy 10 m/s • Calculate the KE of a 500 kg car travelling at a) 10 ms–1 b) 20 ms–1 • • 25 kJ 20 m/s 100 kJ So what effect does doubling the speed have on the kinetic energy? It increases by a factor of 4 How much work does the engine have to do to increase the speed from 10 to 20 ms–1? 75 kJ Work required to increase KE • Ideally, the work done on an object is all transferred to the increase in KE. – e.g. a 40 kg block of ice is pushed across a smooth floor with a force of 100 N for 5 m. what is its final velocity? Work done (F × d) = change in KE (½mv2) so v2 = 2Fd/m = 2×100×5 / 40 So v = 5 m/s • Note that this assumes no energy is lost to friction or air resistance – in real life this is never the case and the final velocity will be smaller. Potential Energy • Potential energy (PE) is the energy stored in an object when you raise it up against the force of gravity. • Energy change = work done So PE gained = work done raising object PE change = force × distance =(mg) × h • So change in PE = mgh Potential energy • A roofer carries 20 kg of tiles up a 10 m ladder to the roof. – What is the gain in PE of the tiles? 2 kJ – How much work did the roofer do lifting them? 2 kJ – He accidentally drops one 500 g tile. How much PE does it lose as it falls to the ground? 50 J Conservation of energy • “Energy cannot be created or destroyed, it can only be transferred from one form to another” • What energy transfers are chemical → kinetic →potential potential → kinetic taking place in the picture? Objects falling due to gravity • When an object falls, PE is converted to KE. – e.g. a 20 kg cannon ball is dropped from the top of the Eiffel tower, which is 320 m high. What is the maximum speed of the ball as it hits the ground? loss of PE (mgh) = gain in KE (½mv2) So v2 = 2mgh/m = 2×20×9.81×320 / 20 so v = 80 ms–1 (1sf) • Why will the actual final speed be less than this? – Wind resistance will limit it PE and KE can interchange • E.g. on a rollercoaster this can happen several times… In fact, work done to overcome friction and air resistance = mgh – ½mv2 loss of PE (mgh) = gain in KE (½mv2) v2 = 2mgh/m = 2gh = 2×9.81×50 Qs on p. 152 so v = 31.6 m/s Power • Power is the rate of transfer of energy – (or the rate of doing work, therefore) E W P t t units: watt (W) 1 W = 1 Js–1 Motive power • When a powered vehicle moves at constant speed: • Work done/s = force × distance/s = Fv • Constant speed so resistive forces = motive force • When a powered vehicle gains speed • Motive power = gain in KE/s + energy lost to resistance/s • Accelerating so motive force > resistive forces Efficiency • No machine which converts energy from one form to another is 100% efficient – Some energy is always “lost” – This is often due to friction of some kind – “wasted” energy (all energy?) tends to end up as heat useful energy transferred by machine Efficiency total energy supplied to machine work done by machine output power energy supplied to machine input power • Now try questions on pp.160–1