* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 4. Transport/pdf (DR)
Atomic theory wikipedia , lookup
Fictitious force wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Classical mechanics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Equations of motion wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Hunting oscillation wikipedia , lookup
Center of mass wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Matter wave wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Faster-than-light wikipedia , lookup
Variable speed of light wikipedia , lookup
Seismometer wikipedia , lookup
Specific impulse wikipedia , lookup
Centripetal force wikipedia , lookup
Speed Speed is a measure of how quickly distance is being covered. Speed= Distance time v= d t v = speed or velocity in metres per second, m/s or ms-1, d = distance in metres, m, t = time in seconds, s. Question Express 18 km/h in m/s. Question Express 18 km/h in m/s. Answer v= d t v= 18,000 3,600 v = 5.0 m/s v=? d = 18 km = 18,000 m t = 1 hour = 3,600 s Average Speed Experiment to find the average speed of a trolley running down a slope. QED • The distance between the light gates is measured using a metre stick. • The time between the light gates is measured using a QED timer. • The average speed is then calculated using the equation: Average Speed = Distance between light gates Time between light gates Instantaneous Speed Experiment to find the instantaneous speed of a trolley at the bottom of a slope. QED • The length of the card is measured using a metre stick. • The time for the card to pass through the light gate is measured using a QED timer. • The instantaneous speed is then calculated using the equation: Instantaneous Speed = Length of card Time to pass through light gate Warning the words length and time on their own are too vague. Acceleration Acceleration is a measure of how quickly velocity is changing. Example 1 Time in seconds 0 1 2 3 Velocity in m/s 0 4 8 12 The velocity is increasing at 4 m/s every second. Example 2 Time in seconds 0 1 2 3 Velocity in m/s 21 25 29 33 The velocity is increasing at 4 m/s per second. Velocity in m/s 4 8 12 16 It is accelerating at 2 m/s per second. Example 3 Time in seconds 0 2 4 6 Example 4 Time in seconds 0 2 4 6 Velocity in m/s 5 15 25 35 It is accelerating at 5 meters per second squared, (general) 5 m/s2, (credit) 5 ms-2, (higher). Equation Acceleration can be found using the equation: v-u a= t a = acceleration in metres per second squared, m/s2. v = final velocity in metres per second, m/s, u = initial velocity in metres per second, m/s, t = time between initial and final velocities in seconds, s. • On a few occasions the units of acceleration may be in mph per second. • The acceleration equation can be rearranged into: v = u + at • Another way of writing the acceleration equation is: a= ∆v ∆t a = acceleration in metres per second squared, m/s2. ∆v = change in velocity in metres per second, m/s, ∆t = change in time in seconds, s. Acceleration Experiment to find the acceleration of a trolley running down a slope. Method 1: Single card with a double beam. QED The QED timer is used to record: • The time for the card to pass through the first light gate (t1). • The time for the card to pass through the second light gate (t2). • The time for the card to move between the light gates (t3). The length of the card is measured using a metre stick. initial velocity (u) = length of card t1 final velocity (v) = length of card t2 acceleration (a) = final velocity (v) - initial velocity (u) time between velocities (t3) Acceleration Experiment to find the acceleration of a trolley running down a slope. Method 2: Double card with a single beam. QED The QED timer is used to record: • The time for the first card to pass through the light gate (t1). • The time for the second card to pass through the light gate (t2). • The time between the first and second cards passing through the light gate (t3). The lengths of the cards is measured using a metre stick. initial velocity (u) = final velocity (v) = acceleration (a) = length of card t1 length of card t2 final velocity (v) - initial velocity (u) time between velocities (t3) Speed-time Graphs v Constant speed. Acceleration = 0 m/s2. t v Speed increasing at a constant rate. Constant acceleration. t Speed-time Graphs v Speed decreasing at a constant rate. Constant deceleration. t v Speed increasing at a increasing rate. Increasing acceleration. t Speed-time Graphs Consider the following speed-time graph. velocity in m/s 10 4 2 1 6 time in seconds Distance travelled = average speed x time d=vxt d= d= Area under the graph = area 1 + Area 2 Area = rectangle + triangle (u + v)t 2 Area = (l x b) + (1/2 x l x b) (4 + 10) x 6 d = 42 m 2 Area = (4 x 6) + (1/2 x 6 x 6) Area = 42 m The area under any speed-time graph is equal to the distance travelled. Speed-time Graphs Example: velocity in m/s 10 2 2 1 3 5 Questions: • describe the motion. • calculate both accelerations. • calculate the distance travelled. 12 time in seconds Answers: For the first 5 seconds the speed is increasing at a constant rate. For the next 7 seconds the speed is decreasing at a constant rate. a=v-u t v = 10 m/s u = 0 m/s t=5s a=? v = 2 m/s u = 10 m/s t=7s a=? 10 - 0 5 a = 2 m/s2 2 - 10 7 a = -1.14 m/s2 acceleration = 2 m/s2 deceleration = 1.14 m/s2 a= a= Speed-time Graphs Distance travelled = area under a speed-time graph. Distance travelled = Area 1 + Area 2 + Area 3 Distance travelled = (1/2 x 5 x 10) + (1/2 x 7 x 8) + (7 x 2) Distance travelled = 25 + 28 + 14 Distance travelled = 67 m. Forces Forces can: • Change the speed of an object, • Change the shape of an object, • Change the direction of a moving object. Forces can be measured using a newton balance. Force The unit of force is the newton (N). Use a small ‘n’ when writing the word, but a large ‘N’ when writing the letter. Weight and Mass Mass is a measure of the quantity of matter that a body has. Mass is measured in kilograms. Weight is a measure of the force of gravity pulling on a mass. Weight is measured in newtons. Mass in kg Weight in N 1 10 2 20 3 30 10 100 From the table we see that the weight of a body is ten times its mass. The ‘ten’ is called the gravitational field strength. Weight = mass x gravitational field strength W = mg W = weight in newtons, N m = mass in kilograms, kg g = gravitational field strength in newtons per kilogram, N/kg. The gravitational field strength on the surface of the Earth is approximately 10 N/kg. Gravitational field strength is different on different planets and decreases as you move away from the surface of a planet. There is nowhere in the universe where the gravitational field strength is 0 N/kg. Weightlessness is a myth. Weight Problem: A girl has a mass of 50 kg on the Earth. Find her mass and weight on the Earth and Moon. The gravitational field strength on the Moon is 1.6 N/kg. Answer: Mass = 50 kg on the Earth and on the Moon. Weight on Earth = mg = 50 x 10 = 500 N Weight on Moon = mg = 50 x 1.6 = 80 N Crash Test Dummy What happens to the passenger when the driver suddenly applies the brakes? With a seat belt. The seat belt applies a force to the passenger and this causes her to slow down with the car (Newton’s 2nd law, F=ma). Crash Test Dummy Without a seat belt. The passenger will obey Newton’s 1st law and will continue to move in a straight line at a constant speed until she hits the windscreen. Forces and Motion Consider a mass being dragged over a frictionless surface. Pulling Force Mass Smooth surface velocity Time The mass will accelerate at a constant rate. Forces and Motion Consider a mass being dragged over a rough surface. Frictional Force Pulling Force Mass Rough surface velocity Constant velocity Increasing velocity Decreasing acceleration Time • At first the mass accelerates because the frictional force is less than the pulling force. • As the mass moves faster the frictional force increases. • Eventually the frictional force will equal the pulling force. The mass will then move at a constant velocity because there is no unbalanced force. Friction Friction can be reduced by: • Polishing the surfaces, • Using oil (in a car engine), • Using rollers (wheels), • Using a cushion of air (hovercraft) • Streamlining the body (car). Sometimes we need friction: • Brakes, • Shoes on road, • Nails in wood. Stopping distance Speed Stopping distance equals the area under the graph Reaction time Time to stop Time Stopping distance Speed Wet roads, worn tyres or icy roads will increase the stopping distance. Reaction time Time to stop Time Stopping distance Speed A faster car has a larger stopping distance. Reaction time Time to stop Time Stopping distance Speed A tired or drunk driver will take longer to react. They will have a larger stopping distance Reaction time Time to stop Time Acceleration and Force Aim: To find relationship between acceleration of a fixed mass and the unbalanced force acting on it. Apparatus: QED Double mask Light gate Vehicle To air blower Linear air track Method: • The acceleration of the vehicle was measured using a light gate connected to a QED computer. • The unbalanced force was taken to be the weight of the hanging masses, and was calculated using W=mg. • The mass of the vehicle was kept constant throughout the experiment. hanging masses Table of Results: Mass in grams Weight in newtons Acceleration 10 0.1 0.29 20 0.2 0.57 30 0.3 0.86 40 0.4 1.14 50 0.5 1.43 60 0.6 1.71 70 0.7 2.00 in m/s2 Acceleration and Force Graph of Results: acceleration in m/s2 X X X X X X X Force in newtons Conclusion: Since we have a straight line graph through the origin we can conclude that the acceleration of a fixed mass is directly proportional to the unbalanced force acting on the mass. aαF -(1) Acceleration and Mass Aim: To find the relationship between the acceleration, produced by a fixed force, and the mass of the object being accelerated. Apparatus: QED Double mask Light gate Vehicle To air blower Linear air track Method: • The acceleration of the vehicle was measured using a light gate connected to a QED computer. • The mass of the vehicle was measured using scales.. • The hanging mass and hence the unbalanced force was kept constant throughout the experiment. hanging masses Table of Results: Mass in kilograms 1/Mass in 1/kg Acceleration 0.3 3.33 2.33 0.7 1.43 1.00 1.0 1.00 0.70 1.4 0.71 0.50 1.7 0.59 0.41 in m/s2 Acceleration and mass Graphs of Results: acceleration in m/s2 acceleration in m/s2 X X X X X X X X X Mass in kilograms X 1/Mass in 1/kg Conclusion: Since the second graph is a straight line graph through the origin we can conclude that the acceleration of a mass by a fixed force is inversely proportional to the mass. aα 1 m -(2) Newton’s Second Law From the first experiment we discovered that aαF -(1), mass constant From the second experiment we discovered that: aα 1 m -(2), unbalanced force constant Combining equations (1) & (2) we get: aα F m Rearranging this becomes: Fαmxa Combining equations (1) & (2) we get: aα F m Rearranging this becomes: Fαmxa A proportion sign (α) can be replaced with a constant (k) and an equals sign. F = kma By defining one newton (N) as the force which will accelerate a mass of one kilogram (kg) at a rate of one meter per second squared (m/s2) the constant will become equal to one and the equation will become: F = ma This equation is known as Newton’s second law. Energy Definition of energy. A body possesses energy if it has the ability to do work. Conservation of energy. Energy cannot be created or destroyed it can only change from one form to another. Forms of energy. Energy Heat, Eh = cm∆T, Eh = lm Sound Light, E = hf Kinetic, Ek =1/2mv2 Potential, Ep = mgh Chemical Electrical, E = ItV Nuclear, E = mc2 Units of energy. Energy is measured in units of: • joules (J), • Kilowatt hours (kWh). Work (Energy) Energy is transferred (work is done) when a force is applied through a distance. Mass (m) Force (F) Distance (d) Obviously the energy transferred to mass m will be directly proportional to: • the size of the force applied, • the distance the mass is moved. Energy Transferred α force - (1) Energy Transferred α distance - (2) Combining equations (1) and (2) we get: Energy Transferred or Work done α force x distance Work done (Ew) α force x distance Ew = k x F x d By defining one joule of energy as the energy transferred when a force of one newton is applied through a distance of one metre the constant becomes equal to one and the equation becomes: Ew = F x d Ew = work done in joules, J F = force in newtons, N d = distance in metres, m Work Done (EW) Example:A man drags a 5kg box a distance of 20m using a 30N force. (i) Calculate the work done by the force. (ii) Suggest what is likely to happen to the work transferred. Answers (i) EW = F x d EW = 20 x 30 EW = 600 joules (ii) Some energy will change to heat energy (the box, air and ground will heat up). Some energy will be changed to sound. Some energy will change to kinetic energy (the box will move faster). Potential Energy A body can be lifted at a constant velocity if an upward force equal to the body’s weight is applied. Mass (m) F h Mass (m) W = mg The potential energy gained by the mass will be equal to the work done by the lifting force. Ew = F x d Ew = Ep F = W = mg d=h Ep = mgh Ep = potential energy in joules, J. m = mass in kilograms, kg. g = gravitational field strength = 10 N/kg on earth. h = height in metres, m. Kinetic Energy time = 0 s Initial velocity = 0 m/s Mass (m) time = t s final velocity = v Force (F) Mass (m) Distance (d) If there are no frictional forces then all the work done will be transferred into kinetic energy. Ew = F x d E w = Ek m(v - u) mv = t t vt (v + u)t d = vt = = 2 2 F = ma = mv Ek = t 2 Ek = mv 2 x vt 2 Ek = kinetic energy in joules, J m = mass in kilograms, kg v = velocity in meters per second, m/s. Relationship between Kinetic Energy and Velocity Aim: To confirm that Kinetic Energy is directly proportional to velocity squared. Apparatus: QED Slotted weights Vehicle Single mask Light gate h To air blower Linear air track hanging masses Method: • The velocity of the vehicle was measured using a light gate connected to a QED computer. • The kinetic energy transferred to the vehicle and masses is equal to the potential energy lost by the hanging masses. (Ek = Ep = mgh). 10g slotted masses were used. g was taken to be 10N/kg. h was measured using a metre stick. • The total mass was kept constant by transferring slotted masses from the vehicle to the hanging masses. Relationship between Kinetic Energy and Velocity Table of Results: Mass in grams Energy in joules Velocity in m/s Velocity2 in (m/s)2 10 0.08 0.48 0.23 20 0.16 0.68 0.46 30 0.24 0.83 0.69 40 0.32 0.96 0.92 50 0.40 1.07 1.15 60 0.48 1.18 1.38 70 0.56 1.27 1.61 80 0.64 1.36 1.84 Graph of Results: velocity2 in (m/s)2 x x x x x x x x EK in joules Conclusion: Since we have a straight line through the origin we can conclude that kinetic energy is directly proportional to the square of its velocity. Ek α v2 Conservation of energy Example A pendulum swings as shown. 2 kg 0.4 m v Find: • the kinetic energy at the bottom of the swing, • the speed at the bottom of the swing. The kinetic at the bottom will equal the potential energy at the top. Ek = 8 J Ek = Ep Ek = mgh Ek = 2 x Ek = 8 J 10 x 0. 4 mv2 = 8 2 v2 = 8 x 2 2 v = 2.83 m/s Conservation of energy A shortcut 2 kg 0.4 m v Ek = Ep mv2 = mgh 2 v2 = 2gh Ek = Ep mv2 = mgh 2 v2 = 2gh v= √2gh The velocity does not depend on the mass. This equation is true when all the potential energy changes to kinetic energy. The velocity at the bottom of the swing in the previous question could have been calculated: √2gh v = √2 x 10 x 0.4 v = √8 v= v = 2.83 m/s Power Power is a measure of how quickly energy is transferred. Example: If 20 joules of energy is transferred in 5 seconds then the power will be 4 joules per second or 4 watts (W). Power can be found using the equation: Power = Energy Time E P= t P = power in watts, W E = energy in joules, J t = time in seconds, s. 1 watt = 1 joule per second. Power of a pupil Title: Power of a pupil. Aim: To find the power of a pupil running up the stairs. Apparatus: Stopclock Pupil Scales h Method: • The height of the stairs was measured using a metre stick. • The time to run up the stairs was measured using a stop clock. • The mass of the pupil was measured using a set of scales. • ‘g’ for the Earth was found in data tables to 10 N/kg. • (If the pupil is running at the same speed at both the bottom and top of the stairs then there will be no gain in kinetic energy.) • The power was calculated using the equation. P= mgh t Results: • height of stairs = 2.5 m. • time to run up the stairs = 2.91 s. • mass of pupil = 60 kg. • gravitational field strength = 10 N/kg (tables). P= mgh t 60 x 10 x 2.4 P= 2.91 P = 495 W Power of a Water Pump Title: Power of a Water Pump. Aim: To find the power output of a water pump. Apparatus: Ignore - siphon effect V Labpack A 1 litre mark h Stopclock Water pump 2.54s Procedure: Measure • The time it takes to pump one litre of water into the bottle. • The height the water is being pumped. • The labpack voltage (should be set at 12 V). • The current drawn from the labpack. Calculate: • Input power to the pump. • Output power of the pump (1 litre of water has a mass of 1kg). • Efficiency of the pump. Some equations Output power = Pout = Potential energy Time mgh t m = 1kg g = 10N/kg Input power = current x voltage Pin = IV Efficiency Efficiency = Power out x 100% Power in P Eff = out x 100% Pin