Work Energy Power Slides [PDF]

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A



UNIT 6



Energy, Work & Power B



¾ show understanding that kinetic energy, elastic potential energy, gravitational potential energy, chemical potential energy and thermal energy are examples of different forms of energy ¾ state the principle of the conservation of energy ¾ apply the principle of the conservation of energy to new situations or to solve related problems ¾ state that kinetic energy Ek = ½ mv2 and gravitational potential energy Ep = mgh (for potential energy changes near the Earth’s surface) ¾ apply the relationships for kinetic energy and potential energy to new situations or to solve related problems ¾ recall and apply the relationship work done = force x distance moved in the direction of the force to new situations or to solve related problems ¾recall and apply the relationship power = work done / time taken to new situations or to solve related problems



Who is doing work ? Work is done only when a force produces motion in the direction of the force. Wall does not moved. ∴ Man is not doing work!!! F



No Motion



F Motion Motion is not in the direction of the force. ∴ Man is not doing work in carrying the parcel!!!



Work is the product of the force on a body and the distance moved in the direction of the force. SI unit for work is joule. Work = Force x Distance moved in the direction of the force W (J)



= F x d (N) (m)



Definition of One Joule: One joule of work is done when a force of moves an object through a distance of one the direction of the force.



one newton meter in



Moment of a force: M = Force x Perpendicular distance from pivot Units: Work



Units:



Nm



= N



m



W



= Force x distance along direction of force



J



= N



m



‡ Moment of force



Eg. 1 An object of weight 5N is pulled a distance of 40 m by a force of 20N and the object moves in the same direction as the force. Calculate the work done by the force. 40m



20N 5N



Solution: Work Done = F x distance moved in the direction of the applied force = 20 N x 40m = 800 Nm = 800 J



More examples…



FORCE



1000N



Distance Moved in the direction of the force 2m



400N



Work Done



1200J 50cm



4000J



Eg. 2 A box of mass 1kg is pulled 50cm along a level floor by a horizontal force 5N. The box is then raised vertically onto a table 80cm high. What is the total work done on the box? (Take g = 10N/kg)



0.8m 0.5m 5N



5N 10N



10N



0.5m 5N



5N 10N Solution: Work Done by horizontal force



Weight of box



10N



= ( 5N ) x (0.50m) = 2.5 J = 1kg x 10 N/kg = 10 N



Minimum force required to raise box vertically = 10 N



Solution: Work Done by vertical force 0.8m



= 10 N x 0.80 m



5N



= 8 J Total work done = 2.5 J + 8 J = 10.5 J



10N



a. Gravitational potential energy (GPE) is the energy a body possesses due to its position relative to the centre of the earth. b. The higher the object, the greater its GPE. Gravitational Potential Energy = mgh (Joule) where: m = mass (kg) h = height of object above reference level (m) g = gravitational acceleration =10 m/s2



Eg. 1 A mass of 40 kg was lifted through a vertical distance of 6 metres in a time of 5 seconds. (Take g = 10N/kg) Calculate a. the weight lifted. b. the potential energy gained by the mass Solution:



a. Weight lifted = mg = 40 kg x 10 N/kg = 400 N 6m b. Gained in GPE = mgh = 40 kg x 10 N/kg x 6 m = 2400 Nm = 2400 J Reference level



Eg. 2 What is the gravitational potential energy of a 400g apple 3 m above the ground? (Take g = 10ms-2) Solution: GPE of apple = mgh = 0.400 kg x 10 N/kg x 3 m 3m



= 12 Nm = 12 J



Eg. 3 A man uses a pulley to move a box along a slope. The box weighs 400N and moves a distance of 2 m along the slope when the man pulls on the taut rope with a force of 220N. a. What is the work done by the man? b. What is the gain in GPE? c. Explain why the work done by the man is larger than the gain in GPE? 220N 2m



1m



220N 2m



1m



Solution: a. Distance moved by the box along the slope when the man pulls on the rope = 2m Work Done by the man = F x d = 220 N x 2 m = 440 J b. State the vertical height moved by the box = 1m



Gain in GPE = mgh = 400 N x 1 m = 400 J c. Explain why the work done by the man is larger than the gain in GPE? Energy is used to overcome friction between the block and the slope.



1. Kinetic energy (KE) is the energy a body posses due to its movement. 2. Any object which is moving has kinetic energy. 3. Motion may be translational or rotational. 4. The faster the object moves, the greater its KE. Translational Kinetic Energy = ½mv2 (Joules) where:



m= mass (kg) v= velocity or speed of object (m/s)



Eg. 1 A small car of mass 1000 kg travelling at 10 m/s. Calculate the a. kinetic energy of the car. b. The car is brought to rest by applying brakes. What was the final kinetic energy of the car? c. Energy is always conserved. Explain why the initial kinetic energy of the car was not equal to the final kinetic energy.



10 m/s



0 m/s



10 m/s



0 m/s



Solution: a. Initial kinetic energy of the car = ½mv2 = ½ x 1000 x (10)2 = 50 000J b. Final kinetic energy of the car = ½mv2 = ½ x 1000 x (0)2 = 0J c. All the kinetic energy has been converted to heat energy (to overcome friction / air resistance)



Eg. 2 A bullet of mass 40g leaves a gun with kinetic energy of 20kJ. Calculate the speed of the bullet. 40g = 0.04kg Solution: K.E.



= ½ m v2



20 000 = 0.5 x 0.04 x v2 20 000 0.5 x 0.04



= v2 v2 = 1000, 000 v = 1000 m/s



Try on your own Q5. Answer: m = 40kg



Eg. 3 The speed-time graph of a car is shown below. The car has a mass of 1200kg. v/kmh-1



a. State the KE of the car at t=0h. K.E. = 0J



A 36



B



b. Find the speed of the car at A in ms-1.



Speed of car at A = 36km / h 36000 = 60 x 60



0 0



= 10 m/s c. Calculate the KE of the car at A.



KE = 60 000J



d. State the KE of the car at B. K.E. =60 kJ



t/h



The law of conservation of energy states that energy cannot be created or destroyed but it can be transformed from one form to another.



Eg. 1 Fig shows an object of mass 1 kg thrown vertically upwards from the ground with an initial velocity of 4 ms-1. It reaches a maximum height of 0.8m before falling back to the ground. a. Find its initial KE. Initial K.E.= ½ m v2 0.8m



1 kg



= ½ x 1 x (4)2 = 8J



b. Fill in the missing information in the table below.



Height



0.8m



1 kg



Remarks



KE



PE



0.80m



0J



8J



Total Energy 8J



0.60m



2J



6J



8J



P.E. > K.E.



0.40m



4J



4J



8J



P.E. = K.E.



0.20m



6J



2J



8J



P.E. < K.E.



0.00m



8J



0J



8J



Minimum P.E. Maximum K.E.



Maximum P.E. Minimum K.E.



Eg. 2 Indicate the type(s) of energy (max/min) at positions A, B, C, D and E. [Refer to textbook Page 106]



A



E B



D



Reference level



C



In the real world, we know that oscillating pendulum will eventually come to a halt. Why?



E.g. 3 A 4.0 Kg mass of metal has fallen through a distance 0.8m onto a horizontal surface in order to test its hardness. a) When the mass has fallen through 0.8m, how much GPE has been transformed? (g = 10N/kg) Solution: GPE = mgh = 4 kg x 10 N/kg x 0.8 m = 32 Nm = 32 J



b) Just before the mass hits the horizontal surface, how much KE does the mass possess? Solution: Applying the Law of Conservation of energy, Gain in KE = Loss in GPE = 32 J



c) What is the velocity of the body just before hitting the horizontal surface? Solution: KE



= ½ m v



32



= ½ (4.0) v2



v2



= 16



v



= 4 m/s



2



d) Assuming that 90% of the energy becomes heat energy, how much heat energy is produced when the body hits the horizontal surface? Solution: Heat energy produced



= 0.9 x 32 = 28.8 J



E.g. A frictionless metal track is curved in a shaped as shown in the diagram. A ball is released from rest at A and slides down the slope. At which pt(s) a. is the KE greatest? b. Same as B? Is it possible for the ball to roll beyond F? Why?



A



1m



F



E



B



D C



1m



We can all complete a cross-country run. But some take 10mins while others take 10 hours.



Power is defined as the rate of doing work or the rate of change of energy from one form to another.



Power = p =



Work done Time taken W t



=



Energy change Time taken



SI unit is W or Js-1



where P is power W is work done t is time taken



(Watts, W) (Joules, J) (sec, s)



E.g. 1



Fill in the blanks



Force



Distance moved in the direction of force



Time taken



3N



2m



4s



6J



1.5W



50N



12m



10s



600J



60W



600N



5m



12s



Work Done



3000J



Power



250W



E.g. 2 A bricklayer lifts 50 bricks, each weighing 15N, through a vertical height of 1.2m in one minute and places them at rest on a wall. Calculate a) the work done, Solution: Work Done = F x d = 50 x 15 N x 1.2m = 900 Nm = 900 J



A bricklayer lifts 50 bricks, each weighing 15N, through a vertical height of 1.2m in one minute and places them at rest on a wall. Calculate b) the average power needed, Solution:



Power = Work Done Time taken 900 J = 1 min = 900 W



Power = Work Done Time taken 900 J = 60 s = 15 W



E.g. 3 A staircase in a building has 75 steps, each one 15cm. A student of weight 500N runs up the staircase in 30s. What is the student’s average power while climbing the stairs? Solution: Height of 1 step = 15cm = 0.15 m Height of 75 steps, h = 75 x 0.15 = 11.25m Work Done in running up 75 steps = F x d = 500N x 11.25m = 5 625 J



Average



Power = Work Done Time taken 5625 J = 30 s = 187.5 W