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Work, Energy, and Power. Work. What is Work and Energy?. Energy(E)- the ability to do work Work (W)- A change in energy (ΔE) or the product of Force and distance (d) W= Fd Work is only being done when components of the applied force are parallel to the displacement

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## Work, Energy, and Power

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**What is Work and Energy?**• Energy(E)- the ability to do work • Work (W)- A change in energy (ΔE) or the product of Force and distance (d) • W=Fd • Work is only being done when components of the applied force are parallel to the displacement • Both are scalar and are measured in joules • 1 joule=kg m2/s2 or 1 joule=N*m**When is work being done?**• If a hold a 30 kg object at a height of 1.5 meters is work being done? • I am exerting energy…. • Work is NOT being done on the OBJECT, work is being done on muscles, but not the object • The distance that the object has moved is zero so no work is done • W=Fd W=F0**Example**• If the same 30 kg object was pushed to the right with an acceleration of 2m/s2 for 2.0 meters, how much work is done on the object? • W=Fd • Remember F=ma • M=30 kg • a=2 m/s2 • d=2.0 m • W=? • F=ma • F=30(2) = 60N • W=Fd • W=60(2) =120 J**Example**• If I were to lift the 30.0 kg object up off the ground to a height of 1.5 m, how much work is done on the object? • When an object is being lifted against gravity, use g for the acceleration (9.8 m/s2) • W=mgd • m=30 kg • g=9.8 m/s2 • d=1.5 m • W=? • W=30(9.8)(1.5) W= 440 J**Clicker Question**• A 3.0 kg pineapple is held 1.2 m above the floor for 15 s. How much work is done on the pineapple? • A)0.25 J • B)54 J • C)35 J • D) 0 J**Clicker Question**• A 10.0 kg pumpkin is moved horizontally 5.00 m at a constant velocity across a level floor using a horizontal force of 3.00 N. How much work is done in moving the pumpkin? • A) 30 J • B) 294 J • C) 15 J • D) 0 J**Example- Honors**• A 50.0 kg banana box is pulled 11.0 m along a level surface by a rope. If the rope makes an angle with the floor of 35oand the tension in the rope is 90.0 N, how much work is done on the box? • Use the force component that is parallel to the displacement! • m=50.0 kg • d=11.0 m • Theta=35 degrees • Fapplied=90.0 N • W=? • W=Fd • F(x)=?**Example**• A 1385 kg car traveling at 61 km/h is brought to a stop while skidding 42 m. What is the work done on the car by frictional forces? • m=1385 kg • vi-=61 km/hr • Vi=17 m/s • Δx=42 m • Vf=0 • W=? • W=Fd • F=? • F=ma**Nonconservative Forces**• The work produced by nonconservative forces are dependent on the paths taken • Friction and air resistance are types of nonconservative forces • A conservative force, such as gravity, are not dependent on the path taken**How does an angle impact work?**If you have a 10.0 kg object, how much work is done on the object if it is lifted 1 meter straight off the ground? On a ramp that is 30 degrees off the ground (A)? 60 degrees(B)? 1.0m A 10 Kg B 10 Kg**How does an angle impact work?**Wf = Ffd Ff = ukFN FN gets larger as the angle gets smaller So… Ff gets smaller, But you still need to Add the Fg So the work Increases with a Higher angle 1.0m A 10 Kg B 10 Kg**Continued**• Work against gravity will not change, however! • If they end at the same height the work against gravity will not change • The amount of force needed will change according to the angle, but the distance will change as well to get to the same height • So the work against gravity is the SAME**Work, Energy, and Power**Potential Energy**Potential Energy**• Energy can either be potential energy or kinetic energy • Potential energy is stored energy • Examples • Chemical • Elastic (bungee cord, trampoline, bow) • Electrical (static charges) • Gravitational potential energy Energy can be converted into different forms by doing work**Gravitational Potential Energy**• Due to an object’s position (height) measured relative to a reference point • Gravitational Potential Energy – Ep(or GPE) • Ep=mgh • m=mass • g=9.8 m/s2 • h=height**Example**• A 15.0 kg textbook is sitting on a 1.20 m tall table. If the book is lifted 0.80 m above the table, how much gravitational potential energy does it have: • With respect of the table? • With respect to the ground?**Clicker Question**• A 1400 kg roller coaster is moved to the top of a track that is 100 m above the lowest part of the track. What is the gravitional potential energy of the coaster? • A) 2000 J • B) 1.4 x 104 • C) 34000 J • D)1.4 x 106 J**Spring/Elastic Potential Energy**• The energy available for use in deformed elastic objects • Rubber bands, springs in trampolines, pole-vault poles, muscles • For springs, the distance compressed or stretched = x • Spring constant (k) depends on stiffness of spring, measured in N/m • Force needed to stretch the spring 1 meter**Example**• When a 2.00 kg mass is attached to a vertical spring, the spring is stretched 10.0 cm such that the mass is 50.0 cm above the table • What is the gravitational potential energy associated with the mass relative to the table? • What is the spring’s elastic potential energy is the spring constant is 400.0 N/m?**Work, Energy and Power**Kinetic Energy**Kinetic Energy**• Energy of motion • scalar • Ek=1/2 m v2 • Ek= kinetic energy • m=mass • v=speed**Example**• A 60.0 kg student is running at a uniform speed of 5.70 m/s. What is the kinetic energy of the student? • Ek=1/2 m v2 • m=60 kg • v=5.70 m/s • Ek=? • Ek=1/2(60)(5.7)2 • Ek=975 J**Clicker Question**• The kinetic energy of a 2.1 kg rotten tomato is 1000 J. How fast is it moving? • A) 15.4 m/s • B) 31 m/s • C) 961 m/s • D) 4000 m/s**Work Energy Theorem**• If a net force is acting on an objet then the object must be accelerating • The change in kinetic energy is proportional to the net force • ΔEk=Fnetd • d=distance**Example**• A sprinter exerts a net force of 260 N over a distance of 35 meters. What is his change in kinetic energy? • ΔEk=Fnetd • Fnet=260 N • d= 35 m • ΔEk=? • ΔEk=260 (35) • ΔEk=9100 J**Example**• A student pushes a 25 kg crate which is initially at rest with a force of 160 N over a distance of 15 meters. If there is 75 N of friction, what is the final speed of the crate? • ΔEk=Fnetd • Ek=1/2mv2 • m=25 kg • F applied=160 N • d=15 m • Ff=75 N**Work, Energy, and Power**Conservation of Energy**Law of Conservation of Energy**• Energy cannot be created or destroyed, only converted into other forms of energy • TOTAL energy is always conserved • Potential energy can be converted to kinetic energy as an object moves • When only conservative forces act on object potential energy is completely converted to kinetic energy • When nonconservative forces like friction act on an object, some energy will be converted to heat**Mechanical Energy**• Mechanical Energy is the sum of kinetic energy and all forms of potential energy associated with an object • ME=KE + PE • When only conservative forces act on an object then mechanical energy is conserved**Law of Conservation of Energy (Quantitatively)**• Initial Energy= Final Energy • Ei = Ef • GPEi + Kei = GPEf + Kef • mghi + ½ mv2i = mghf + ½ mv2f • GPE=gravitational potential energy**Example**• A student falls from the building, if they reach the ground at 5.0 m/s , what height did they fall from? • Vf=5.0 m/s • Hi=? • Vi=0 • Hf=0 • GPEi + KEi= GPEf + KEf**Example**• While jumping over The Great Wall of China an 82 kg skateboarder is needs to leave the ramp traveling at 22 m/s.A) How much potential energy is needed to jump over? B) What minimum height should the ramp be? • m=82 kg • Vf=22 m/s • g=9.8 m/s2 • GPE=? • h=? • Ei=Ef • GPEi + KEi= GPEf + KEf • GPEi =Kef • GPE = ½ mv2f**Clicker Question**• A 66 kg skateboarder jumps The Great Wall of China, clearly. At the peak of jump he is 18 m high and traveling at 12 m/s . Assuming he started at rest, find his initial height. • A) 10 m • B) 19 m • C) 25 m • D) 30 m**Clicker Question**• A 75 kg snowboarder slides up a frictionless rail to a height of 1.75m and slides across it at 2.50 m/s. How much kinetic energy did he have before he went up the rail? • A) 1520 J • B) 1380 J • C) 200 J • D) Impossible to solve**Roller Coasters**• Although not perfectly energy efficient, they are a fun way to view how work, gravitational potential and kinetic energy are exchanged**The Downhill skier**• When a nonconservative force is applied (friction) the work is negative because it is removing energy from the system What’s this?**Work, Energy, and Power**Power**Power**• Power (P) is the rate of doing work • Measured in J/s or Watts (W) • Power= Work/time • P=W/t or P=ΔE/t**Clicker Question**• Mike performed 5 J of work in 10 seconds. Joe did 3 J of work in 5 seconds. Who produced the greater power? • A) Mike • B) Joe • C) Both produced the same amount of power**Example**• Lover’s Leap is a 122 m vertical climb. The record time of 4 min 25 s was achieved by Dan Osman (65 kg). What was his average power output during the climb? • h=122 m • t=4 min 25 s 265 s • m=65 kg • g=9.8 m/s2 • P=? • P=W/t or P= ΔE/t**Example**• A 1.00x103 kg car accelerates from rest to a velocity of 15.0 m/s in 4.00 s. Calculate the power output of the car. Ignore friction. • m=1.00 x 103 kg • Vf=15 m/s • t=4.00 s • P=? • P=W/t or P=ΔE/t**Clicker Question**• A 68 kg student runs up a flight of stairs 3.2 m high in 4.8 seconds. Determine their power output while running up the stairs. • A) 217.6 W • B) 45.33 W • C) 440 W • D) There is no work in this problem, not enough info**Clicker Question**• A 642 kg formula 1 car can reach a speed of 27.78 m/s in 1.7 seconds. What is the power output of the car during this acceleration? • A) 300,000 W • B) 5000 W • C) 150,000 W • D) 3000 W**Another Useful Formula**• P=W/t • P=Fd/t • V=d/t • So.. • P=FV • Note this formula is only useful when the velocity is held constant**Example**• A student uses 140 N to push a block up a ramp at a constant velocity of 2.2 m/s. What is their power output? • F=140 N • V=2.2 m/s • P=? • P=FV • P=140 (2.2) • P=310 W**Clicker Question**• An elevator motor has a power rating of 110 kW. How much force would it exert if it was lifting a load at a constant velocity of 3.0 m/s? • A) 3700 N • B) 37 N • C) 330 N • D) 4 N**Work, Energy, Power**Efficiency**Efficiency**• A measure of how much of the energy that goes into a machine actually gets used • Machines are useful because they allow us to use less force over a longer distance to do the same work**Efficiency of a Machine**• Eff= W out x 100 W in • Eff= P out x 100 P in • There are no units for efficiency because it is a percentage

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