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General Physics 101 PHYS

General Physics 101 PHYS. Dr. Zyad Ahmed Tawfik. Email : zmohammed@inaya.edu.sa. Website : zyadinaya.wordpress.com. Work, energy , and power 6-1 Work 6-2 Kinetic energy 6-3 Potential energy and conservative forces 6-4 Dissipative forces 6-9 power. 6-1 Work. W = F. d.

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General Physics 101 PHYS

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  1. General Physics101 PHYS Dr. Zyad Ahmed Tawfik Email : zmohammed@inaya.edu.sa Website : zyadinaya.wordpress.com

  2. Work, energy , and power 6-1 Work 6-2 Kinetic energy 6-3 Potential energy and conservative forces 6-4 Dissipative forces 6-9 power

  3. 6-1 Work W = F.d units: N.m or joule, j • W = Fd cos q • Work = the product of force and displacement, times the cosine of the angle between the force and the direction of the displacement. • How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m? • W = F d cos q • W = 15 N (12 m) (cos 20) 15 N 20o = 169J

  4. Sign of work: +vework: Force acts in the same direction as the displacement.  object gains energy ve work: Force acts in the opposite direction as the displacement.  object loses energy When force F is at right angles to displacement s (F ⊥s):  cos 90 = 0 • No work is done if: • F = 0 or • Δx = 0 or • q= 90o  F Δxcos = 0  No work is done on the load Norah Ali Al Moneef king Saud university

  5. Example: When Work is Zero • A man carries a bucket of water horizontally at constant velocity. • The force does no work on the bucket • Displacement is horizontal • Force is vertical • cos 90° = 0

  6. example

  7. example • An Eskimo returning pulls a sled as shown. The total mass of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30° and he pulls the sled 5.0 m ? • Suppose µk = 0.200, How much work done on the sled by friction, and the net work if θ = 30° and he pulls the sled 5.0 m ?

  8. Example : A horizontal force F pulls a 10 kg carton across the floor at constant speed. If the coefficient of sliding friction between the carton and the floor is 0.30, how much work is done by F in moving the carton by 5m? The carton moves with constant speed. Thus, the carton is in horizontal equilibrium.  F = f = μk N = μk mg.  Thus F = 0.3 x 10 x 9.8 = 29.4 N  Therefore work done W = FS =(29.4 Cos 0o) =147 J

  9. 6-2 Kinetic Energy Same units as work Remember the Eq. of motion Multiply both sides by m, W = F Dx W = DK Work-Energy Theorem • When work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. • The speed of the system increases if the work done on it is positive( the kinetic energy of the particle increases). • The speed of the system decreases if the net work done on it is negative(the kinetic energy of the object decreases).

  10. An object’s kinetic energy depends on its mass and its speed. The faster something is moving and the heavier it is, the more work it can do, so … If ‘v’ tripled then KE will increase by 9 times. If ‘v’ quadruples, then KE will increase by 16 times example M= 4 kg Norah Ali Al - moneef

  11. Do changes in velocity and mass have the same effect on kinetic energy? • No—changing the velocity of an object will have a greater effect on its kinetic energy than changing its mass. • This is because velocity is squared in the kinetic energy equation. • For instance, doubling the mass of an object will double its kinetic energy. • But doubling its velocity will quadruple its kinetic energy.

  12. example • A 1500 kg car moves down the freeway at 30 m/s • K.E. = ½(1500 kg)(30 m/s)2= 675,000 kg·m2/s2 = 675 kJ • A 2 kg fish jumps out of the water with a speed of 1 m/s • K.E. = ½(2 kg)(1 m/s)2= 1 kg·m2/s2 = 1 J • Calculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cm. • work = force x distance • the child must exert an upward force equal to its weight • the distance moved upwards equals (12 x 20cm) = 2.4m • work = 300 N x 2.4 m= 720 J

  13. Kinetic Energy – the energy of motion • KE = ½ m v 2 • units: kg (m/s) 2 = (kgm/s2) m • = Nm = joule • Work Energy Theorem: W = D KE • Work = the change in kinetic energy of a system. Note: v = speed NOT velocity. The direction of motion has no relevance to kinetic energy. Example : A 6 kg bowling ball moves at 4 m/s.a) How much kinetic energy does the bowling ball have?b) How fast must a 2.5 kg tennis ball move in order to have the same kinetic energy as the bowling ball? K = ½ mb vb² K = ½ (6 kg) (4 m/s)² KE = 48 J KE = ½ mt vt² v = √2 K/m = 6.20 m/s

  14. M=6.00 kg Another solution

  15. Example : A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m. W =F d cos 0 =12x 3x1 = 36J Δk = w 0.5m V2 =W =36 J

  16. Example What acceleration is required to stop a 1000kg car traveling 28 m/s in a distance of 100 meters?

  17. example example

  18. example Calculate the speed of a car of mass 1200kg if its kinetic energy is 15 000J K = ½ m v2 15 000J = ½ x 1200kg x v2 15 000 = 600 x v2 15 000 ÷ 600 = v2 25 = v2 v = 25 speed = 5.0 ms-1

  19. Calculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN. k.E. of car = ½ m v2 = ½ x 900kg x (20ms-1)2 = ½ x 900 x 400 = 450 x 400 k = 180 000 J The work done by the brakes will be equal to this kinetic energy. W = F Δx 180 000 J = 3 kN x Δx 180 000 = 3000 x Δx Δx = 180 000 / 3000 braking distance = 60 m example

  20. 6-3 Potential energy and conservative forces • An object does not have to be moving to have energy. • Some objects have stored energy as a result of their positions or shapes. • When you lift a book up to your desk from the floor or compress a spring to wind a toy, you transfer energy to it. • The energy you transfer is stored, or held in readiness. • It might be used later when the book falls to the floor • Stored energy that results from the position or shape of an object is called potential energy. • This type of energy has the potential to do work.

  21. Gravitational Potential Energy • Potential energy related to an object’s height is called gravitational potential energy. • The gravitational potential energy of an object is equal to the work done to lift it. • Remember that Work = Force × Distance. • The force you use to lift the object is equal to its weight. • The distance you move the object is its height. • You can calculate an object’s gravitational potential energy using this formula: • Gravitational Potential Energy = Weight x Height change in Gravitational potential energy . = mass x gravitational field strength x change in height ΔU = m g Δh

  22. Gravitational potential energy: • Potential energy only depends on y (height) and not on x (lateral distance) • MUST pick a point where potential energy is considered zero!

  23. example Calculate the change in Gravitational potential energy (g.p.e). when a mass of 200 g is lifted upwards by 30 cm. (g = 9.8 Nkg-1) ΔU = m g Δh = 200 g x 9.8 Nkg-1 x 30 cm = 0.200 kg x 9.8 Nkg-1 x 0.30 m = 0.59 J change in g.p.e. = 0.59 J example What is the potential energy of a 12 kg mass raised from the ground to a to a height of 25 m? Potential Energy = weight x height change Weight = 12 x 10 = 120 N Height change = height at end - height at start = 25 - 0 = 25 m Potential energy = 120 x 25 = 3000 J

  24. example Calculate the gravitational potential energy in the following systems a. a car with a mass of 1200 kg at the top of a 42 m high hill (1200 kg)( 9.8m/s/s)(42 m) = 4.9 x 105 b. a 65 kg climber on top of Mt. Everest (8800 m high) (65 kg) (9.8m/s/s) (8800 m) = 5.6 x 106 J c. a 0.52 kg bird flying at an altitude of 550 m (.52 kg) (9.8m/s/s)(550) = 2.8 x 103 J

  25. Norah Ali Al - moneef

  26. example Who is going faster at the bottom? • Assume no friction • Assume both have the same speed pushing off at the top same

  27. Conservation of Energy • Law of conservation of energy - energy cannot be created or destroyed closed system: - all energy remains in the system - nothing can enter or leave open system: - energy present at the beginning of the system may not be at present at the end

  28. Conservation of Energy • When we say that something is conserved it means that it remains constant, it doesn’t mean that the quantity can not change form during that time, but it will always have the same amount. • Conservation of Mechanical Energy: MEi = MEf • initial mechanical energy = final mechanical energy If the only force acting on an object is the force of gravity: Ko + Uo = K + U ½ mvo² + mgho = ½ mv² + mgh

  29. example • At a construction site a 1.50 kg brick is dropped from rest and hit the ground at a speed of 26.0 m/s. Assuming air resistance can be ignored, calculate the gravitational potential energy of the brick before it was dropped? • K+U = Ko+Uo • ½ mv2+mgh=½ mv2o+ mgho • ½ mv2+0 = =o+ mgho • U0 = mgho = ½ (1.50 kg) (26.0m/s)2 = 507 J

  30. A child of mass 40 kg climbs up a wall of height 2.0 m and then steps off. Assuming no significant air resistance calculate the maximum: (a) gravitational potential energy (gpe) of the child (b) speed of the child g = 9.8 Nkg-1 (a)maximum gravitational potential energy ( max gpe) occurs when the child is on the wall Δ U = mgΔh = 40 x 9.8 x 2.0 max gpe = 784 J (b) max speed occurs when the child reaches the ground ½ m v2= m g Δh ½ m v2= 784 J v2= (2 x 784) / 40 v2= 39.2 v = 39.2 max speed = 6.3 ms-1 example

  31. example A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distance. m = 10.0 kg vi = 0 Fp = 250.0 N Fk = 50.0 N Δx = 10.0 m FN A. How much work is done by each force on the cart? Fk Fp Wg = 0 WN = 0 Fg Wp = FpΔx cos = (250.0 N)(10.0 m)cos 0 Wp = 2500 J Wk = FkΔx cos= (50.0 N)(10.0 m)cos 180= -500 J

  32. B. How much kinetic energy has the cart gained? Wnet = ∆KE Wp + Wk = KEf - KEi 2500 J + -500 J = KEf - 0 KEf = 2000J C. What is the carts final speed? KE = 1/2 m v2 v = √((2KE)/(m)) v = √((2(2000 J))/(10.0 kg)) v = 20 m/s

  33. example At B At C

  34. A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a force of 31 kN over a displacement of 2m.Find the upward speed of truck after its displacement.

  35. 6-4 Dissipative forces Non conservative Forces • A non conservative force does not satisfy the conditions of conservative forces • Non conservative forces acting in a system cause a change in the mechanical energy of the system • The work done against friction is greater along the brown path than along the blue path • Because the work done depends on the path, friction is a non conservative force

  36. Work and Energy • If a force (other than gravity) acts on the system and does work • Need Work-Energy relation • or

  37. example On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does it travel if the coefficient of kinetic friction between the sled and the ice is .10? m = 10.0 kg vi = 2.2 m/s vf = 0 m/s µk = .10 W = ∆KE FΔx= Kf - Ki µk = Fk / FN Fk = µk (mg) µk (-mg) Δx = 1/2 m vf2 - 1/2 m vi2 µk (-mg) Δx = - 1/2 m vi2 Δx = (- 1/2 m vi2) / µk (-mg) Δx = (- 1/2 (10.0 kg) (2.2 m/s)2) / (.10 (-10.0 kg) (9.8 m/s2)) Δx = 2.47 m

  38. Example A 20-g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s. W =Δ K F Δ x cosθ = K – K0 F Δ x cosθ= ½ mv² - ½ mvo² F (0.06 m) cos 1800 = 0 - 0 .5 (0.02 kg)(80 m/s)2 F (0.06 m)(-1) = -64 J F = 1067 N Work to stop bullet = change in K.E. for bullet W = ½ mv² - ½ mvo² = 0

  39. example A bus slams on brakes to avoid an accident. The tread marks of the tires are 80 m long. If μk = 0.7, what was the speed before applying brakes? Work = F Δx (cos θ) f = μk.N = μk mg Work = - μk mg Δx ΔK = 1/2 mv2 - ½ mvo2 W = ΔK vo = (2μk g Δ x) ½

  40. correct Example Suppose the initial kinetic and potential energies of a system are 75J and 250J respectively, and that the final kinetic and potential energies of the same system are 300J and -25J respectively. How much work was done on the system by non-conservative forces? 1. 0J 2. 50J 3. -50J 4. 225J 5. -225J Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies. W = (300-75) + ((-25) - 250) = 225 - 275 = -50J.

  41. FN V T correct correct correct W example You are towing a car up a hill with constant velocity. A )The work done on the car by the normal force is: 1. positive2. negative3. zero The normal force is perpendicular to the displacement, hence, does no work. b )The work done on the car by the gravitational force is: 1. positive2. negative3. zero With the surface defined as the x-axis, the x component of gravity is in the opposite direction of the displacement, therefore work is negative. C ) The work done on the car by the tension force is: 1. positive2. negative3. zero Tension is in the same direction as the displacement.

  42. 6-9 power • Power is the rate at which work is done. • Power is the rate of energy transfer by any method. • The SI unit of power is the watt, W • 1 watt = 1 Joule/s • 1000 watts = 1 kW Work= force · distance Power = force·distance/time Power = force·velocity P = F·v

  43. example Calculate the power of a car engine that exerts a force of 40 kN over a distance of 20 m for 10 seconds. W = FΔx = 40 kN x 20 m = 40 000 x 20 m = 800 000 J P = ΔW / Δt = 800 000 J / 10 s power = 80 000 W example • In 1935, R. Goddard, the America rocket pioneer, launched several A-series sounding rockets. Given that the engine had a constant thrust of 890 N, how much power did it transfer to the rocket while traveling at its maximum speed of 113.9 m/s. • P = Fv = 890N x 313.9 m/s = 279 kW.

  44. example • Together, two students exert a force of 825 N in pushing a car a distance of 35 m. • A. How much work do they do on the car? • W = F Δx = 825N (35m) = • B. If the force was doubled, how much work would they do pushing the car the same distance? • W = 2F Δx = 2(825N)(35m) =

  45. Power Delivered by an Elevator Motor • A 1000-kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s? Since the speed is constant, so a = 0 where M = total mass of the system M= 1000+800=1800 Kg

  46. example OR

  47. example In a rifle barrel, a 15.0 g bullet is accelerated from rest to a speed of 780 m/s. a)Find the work that is done on the bullet.? Using Work-Kinetic Energy Theorem, W = ΔKE W = KEf - KEiW = 1/2(0.015 kg)(780 m/s)2 - 0 W = 4.56 kJ b) If the rifle barrel is 72.0 cm long, find the magnitude of the average total force that acted on the bullet. Using W = Fcosθd, F = (4.56 kJ) / (0.72 m) F = 6.33 kN

  48. example • A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. • How much work does the climber do on the backpack? • W = Fd = m g Δx= 7.5kg(9.8m/s2)(8.2m) • b.How much power does the climber expend in this effort? • P = W = 603 J = 0.34 watt • t 1800 s

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