[Music] hey everyone welcome to PL Academy hope you found this video useful if you did it would be awesome if you could subscribe like the video share it with your friends and maybe drop a comment your support really helps me make more videos in this video I covered everything on the syllabus like you can see in the figure work done is the product of force and displacement moved in the direction of the force it is a scalar quantity and its unit is Jewel the formula of the work done is W equals FD where W is work done in Jewel f is the force in Newtons and D is the displacement moved in the direction of the force in meters when the force and the displac M are in the same directions in this case the work done is positive if a 5 Newtons Force acts on a box causing it to move 10 m in the same direction are shown the work done can be calculated using W = FD substituting F = 5 Newtons and D = 10 m we get the work done equals 50 Jewel when the force and the displacement are in the opposite directions in this case the work done is negative if a 5 Newtons Force acts on a box to the left while the Box moves 10 m to the right are shown the work done can be calculated using wal FD substituting F = 5 Newtons and D = -10 M we get the work done equals -50 Jew the negative sign indicates that the Force opposes the displacement when the force F acts on a box at an angle Theta to the horizontal causing the box to move D meters to the right are shown the force F can be resolved into horizontal and vertical component the horizontal component is f cos Theta and the vertical component is f sin Theta F cos Theta is in the same direction to the displacement D this component does work W on the box as F cos Theta * d f sin Theta is perpendicular to the displacement D this component does not work on the box when a person walks 10 m while holding a box weighing 10 Newtons the person exerts a force of 10 Newtons to hold the Box however the Box moves forward in a direction perpendicular to this Force are shown therefore the work done by the person holding the box is zero when the gas in the cylinder does work on a piston of cross-sectional area a the pressure p on the Piston has a constant value the force F due to the gas pressure acting on the Piston causes the Piston to move outward with distance Delta X the work done W by the gas is given by w = f Delta X the force F exerted by the gas is equal to to the pressure p multiplied by the area a as FAL PA a since a Delta X is the change in the volume of the cylinder Delta V therefore the work done by the gas can be expressed as W equals PTA V exam style question one a team of nine dogs can pull a sledge with a combined force of 800 Newtons at a speed of 1.5 5 m/s for 360 minutes what is the average work done by each dog during this time the work done can be calculated using W equal FD the force F of nine dogs equals 800 Newtons the distance D can be calculated using D equals speed V multiplied by the time T substituting V = 1.5 m/s and t equal 360 * 60 seconds so the distance D equals 32,400 m substituting F = 800 Newtons and D = 32,400 M we get the work done by nine dogs equals 25,920 th000 Jewel so the work done by one dog equals 25,920 th000 / by 9 is equal to 2.9 * 10^ 6 Jew for two significant figures exam style question two a piston in a gas supply pump has an area of 600 square cm and it moves a distance of 40 cm during one stroke the pump moves the gas against a fixed pressure of 5,000 pascals how much work is done by the Piston during one stroke the work done by the gas can be calculated using W = FD the force F by the gas can be calculated using F equal p a the pressure p by the gas equals 5,000 pascals the area a equal 600 square cm we convert the area a into square m like this which is 0.06 s m the distance D equal 0.4 m M substituting P = 5,000 a = 0.06 and D = 0.4 we get the work done by the gas W equals 120 Jew energy is the ability to do work it is a scalar quantity and its unit is Jewel there are many form of the energy as follows kinetic energy is the energy due to the movement of a mass gravitational potential energy store in a mass due to change in its position within a uniform gravitational field strength elastic potential energy or stain energy store in an object due to change in its shape chemical potential energy electrical potential energy is the work done by the electric force due to the electric field thermal or heat energy radiation energy for example light infrared or gamma Sound Energy nuclear energy principle of conservation of energy the energy cannot be created and destroyed but can only be transferred from one form to another principle of work energy work done and the energy principal state that mechanical or electrical work done is equal to the energy transferred derive the formula of the kinetic energy as EK equal half of m² v a force F pushes a box along a smooth surface causing a displacement of s the Box's velocity increases from an initial value of U to a final value of V since the surface is smooth there is no thermal energy loss therefore all the work done W by the force f is transferred to an increase in kinetic energy Delta EK K the work done by the force f w equal FS so Delta EK k equal FS the force F equal m a from the equation V ^2 = U ^2 + 2 a s so a s = half of V ^2 - U ^2 substituting a s this equation like this so Delta e k = M2 of V ^2 - U ^2 assuming that the initial velocity U equals 0 therefore Delta EK k equal half of m² V derive the formula of the gravitational potential energy as EP equal mg Delta h a box with a mass m kilg being lifted from the ground to a higher level with the height Delta H since we use the force F that equals Box's weight to hold the box so f equals mg therefore all the work done W by the force f equals mg is transferred to an increase in gravitational potential energy Delta ep the work done by the force f w equal FS so Delta EP equals FS the force f equals mg the distance s equal Delta H therefore Delta = mg Delta H exam style question three when a horizontal Force f is applied to a frictionless trolley over distance s the kinetic energy of the trolley changes from 4 Jew to 8 Jewel if a force of 2f is applied to the trolley over a distance of 2s what will the original kinetic energy of 4 Jewel become the frictionless surface means that no heat energy loss or no work done against friction so the work done by the force f equals the kinetic energy gain the work done by the force f = f * s and EK gain equal 8 J - 4 Jew so FS equals 4 Jew if a force of 2f is applied to to the trotle over a distance of 2s the original kinetic energy of 4 Jewel become the final kinetic energy EK so 4 FS = EK - 4 JW and EK = 4 FS + 4 substituting FS equal 4 we get the final EK equal 20 Jew exam style question 4 a constant force of 9.0 k parallel to an inclined plane moves a body of weight 20 kons through a distance of 40 m along the plane at constant speed the body gains 12 m in height as shown how much of the work done is dissipated as heat the Box moves at constant speed to indicate that its kinetic energy Remains the Same so the work done by 9 Kon Force equals the gravitational potential energy gain Plus work done against friction or dissipated heat energy therefore f * s = mg Delta h plus dissipated heat substituting F = 9,000 Newtons mg = 20,000 Newtons Delta H = 12 M we solve the dissipated heat equals 120,000 Jew exam style question five a uniform solid cuboid of concrete of Dimensions 0.5 M * 1.2 M * 0.4 M and weight 4,000 Newtons rests on a flat surface with the 1.2 M Edge vertical as shown in diagram 1 what is the minimum energy required to roll the cuboid through 90° to the position shown in diagram 2 with the 0.5 M Edge vertical the solid cuboid will topple over over if a vertical line drawn downward from its center of gravity falls outside its base area are shown so the minimum energy required to rise the cuboids center of gravity is mg Delta H therefore Delta H equals the difference between the center of gravity A and B this distance is half of 12 m equals 06 M the distance X is half of the diagonal of cuboid this distance is half of 0.5 m equal 0.25 M the distance X can be calculated using the Pythagorean theorem like this we solve the distance x = 0.65 M so Delta H = 0.65 - 0.6 is equal to 0.05 M substituting mg = 4,000 Newtons and Delta H equal 0.05 M we get the minimum energy required equals 200 juwel exam style question six a box of weight 200 Newtons is pushed so that it moves at a steady speed along a ramp through a height of 1.5 M the ramp makes an angle of 30° with the ground the friction force on the box is 150 Newtons while the box is moving what is the work done by the person the Box moves at constant speed to indicate that its kinetic energy Remains the Same so the work done by the person equals the gravitational potential energy gain Plus work done against friction or dissipated heat energy therefore the work done by the person equals mg Delta H + friction time distance D therefore distance D is the distance along the ramp like this so sin 30 equal Delta h / D we solve the distance D = 1.5 / sin 30 = 3 m substituting mg = 200 Newtons Delta H = 1.5 M friction equal 150 Newtons and D equal 3 m we get the work done by the person equals 750 Jewel exam style question seven a trolley runs from P to Q along a track at Q its potential energy is 50 KJ less than at P at P the kinetic energy of the trolley is 5 KJ between p and Q the work the trolley does against friction is 10 kilj what is the kinetic energy of the trolley at Q the kinetic energy of the trolley at P equal 5 K when the trolley runs from P to Q the work done against friction or heat energy loss is 10 kle and the kinetic energy at Q is EK K therefore the gravitational potential energy loss from P to Q equals 50 k so the gravitational potential energy loss equals the kinetic energy gain Plus work done against friction substituting EP loss equals 50 KJ EK gain equals EK - 5 K and work done against friction equals 10 K we solve the EK at qal 45 KJ power is the work done per unit time or energy transferred per unit time time it is a scalar quantity and its unit is Watts or Jew per second the formula of the power P equal W / T or E / T where p is the power in watts w is work done in Jewel T is time in seconds and E is the energy in Jewel if we substitute work done W equal force f * displacement D into p = w / t where D / T is speed V so the power P equals f * V like this exam style question 8 an area of land is an average of 2 M below sea level to prevent flooding pumps are used to lift rainwater up to sea level what is the minimum pump output power required to deal with 1.3 * 10^ 2 kg of rain per day the power can be calculated using P equal e / T where the energy E equals gravitational potential energy mg Delta H the mass m equal 1.3 * 10^ 9 kg the gravitational field strength g = 9.81 m per squs the Delta h = 2 m the time T is 1 day converting into second which is 86,400 seconds substituting m g Delta H and T in the equation like this we get the power P equals 300 kilow for two significant figures exam style question n the diagram shows a lift system in which the elevator Mass M1 is partly counterbalanced by a heavy weight Mass M2 at what rate does the motor provide energy to the system when the elevator is rising at a steady speed v g equals acceleration of Free Fall the system moves at steady speed to indicate that its kinetic energy Remains the Same so the power input by the motor equals the rate of gravitational potential of the system loss when the elevator moves Upward at steady speed V with distance Delta H so its kinetic energy is constant and its gravitational potential energy gain as M1 G Delta H the mass M2 moves downward at steady speed V with distance Delta H so its kinetic energy is constant and its gravitational potential energy loss as m2g Delta H so the power input by the motor equals m2g Delta hus M1 G Delta h / T we factor out G Delta Delta H like this the Delta H over T is the steady speed V so the power input by the motor equals M2 minus M1 GV exam style question 10 a mass of 2 kg rests on a frictionless surface it is attached to a 1 kg mass by a light thin string which passes over a frictionless pul the 1 kg mass is released and it accelerates downwards what is the speed of the 2 kg mass as the 1 kg Mass hits the floor having fallen a distance of 0.5 M this question can solve using the suat equations or the principle of conservation of energy but I will solve using the principle of conservation of energy when a 1 kg Mass moves downward 0.5 m causing its speed to increase from 0 to V this causes a 2 kg Mass also moves 0.5 m to the right and its speed also increases from 0er to V the frictionless surface causes no heat loss or no work done against friction therefore the gravitational potential energy of system loss equals the kinetic energy of system gain the EP loss equals mg Delta H and EK gain equal half m² V - half m² U A 1 kg Mass moves downward causes EP loss equal 1 * 9.81 * 0.5 both masses increase its speed from 0 to V to cause EK gain equal half of 1 + 2 * Square V - 0 we solve the speed V equal 1.8 m/s for two significant figures exam style question 11 an escalator is 60 M long and lifts passengers through a vertical height of 30 m as shown to drive the escalator against the forces of friction when there are no passengers requires a power of 2 kilow the escalator is used by passengers of average mass 60 kg and the power to overcome friction remains constant how much power is required to drive the escalator when it is carrying 20 passengers and is traveling at 0.75 m/ seconds the escalator moves at constant speed 0.75 m/s to indicate that its kinetic energy Remains the Same therefore the power drive the escalator and passengers equals power when no passenger plus power drives 20 passengers the the power drives 20 passengers equals work done FD divided by time t f equals the total weight of 20 passengers equal 60 * 9.81 * 20 is equal to 11,720 newtons D equals the vertical height of the escalator equal 30 m t equals the time of the escalator moving at 0.75 m/s for 60 m so T = 60 / 0.75 is equal to 80 Seconds substituting f = 11,720 d = 30 and T = 80 we get the power to drive 20 passengers equals 44145 wats therefore the power drives the escalator and passengers equals 2,000 + 4,44 14.5 we get the results is 6.4 kilow exam style question 12 a conveyor belt is driven at velocity V by a motor sand drops vertically onto the belt at a rate of M kg/s what is the additional power needed to keep the conveyor belt moving at a steady speed V when the sand starts to fall on it when sand drops vertically onto the belt as shown the belt exerts a force F on the sand and the sand exerts an equal and opposite Force F on the Belt the additional power needed to keep the conveyor belt moving at a steady speed V to calculate using p equals f * V the force F can be calculated using mvus mu U / T the initial velocity U of the sand equals zero the final velocity V of the sand equals V the sound's mass m equal m and the time T = 1 second substituting all quantities in the equation like this we get the force FAL MV substituting FAL MV and V equal V we get the additional power equal m² V efficiency is the ratio of the useful energy or work or power output from the system to the total energy or work or power input exam style question 13 a constant force F acting on a car of mass m moves the car up the slope through a distance s at constant velocity V the angle of the slope to the horizontal is Alpha which expression gives the efficiency of the process the efficiency of this question can be calculated using the efficiency equals useful energy or work output divided by total energy or work input the useful energy output is the gravitational potential energy gain of the car equals mg Delta H the Delta H is the vertical height and sin Alpha equal Delta h / s so Delta h = s * sin Alpha therefore EP gain equals MGS sin Alpha total energy input equals the work done by the force F so the work done by the force f equals f * s therefore the efficiency equals MGS sin Alpha / FS we cancel out s like this since the result is mg sin Alpha over F exam style question 14 a bow of mass 400 G shoots an arrow of mass 120 g vertically upwards the potential energy stored in the bow just before release is 80 Jewel the system has an efficiency of 28% what is the height reached by the Arrow when air resistance is neglected the efficiency can be calculated using the efficiency equals useful energy output divided by total energy input the efficiency equals 28% the useful energy output equals GP gain equals mg Delta H the total energy input equals the potential energy storing in the bow GP gain equals 0.12 kg * 9.81 * Delta H the potential energy storing in the bow equals 80 Jewel we solve the Delta H equal 0 M for two significant figures exam style question 15 water from a reservoir is fed to the turbine of a hydroelectric system at a rate of 500 kg/s the reservoir is 300 m above the level of the turbine the electrical output from the generator driven by the turbine is 200 amp at a potential difference of 6,000 volts what is the efficiency of the system the efficiency of this question can be calculated using the efficiency equals useful power output divided by total power input the useful power output equals the electrical power equals VI the total power input equals the gravitational potential energy per second of the water the electrical power equals 6,000 volts time 200 amp the gravitational potential energy per second of the water equal 500 kg/s * 9.81 * 300 M therefore the efficiency equals 82% for two significant figures I hope you found this video helpful if you did I would be grateful if you would subscribe share like and leave a positive comment your support will encourage me to create more 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