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Work, Energy, Power. What’s the difference?. Force is the agent of change Energy is a measure of change Work is a way of transferring energy from one system to another. What is work?. Work= force*displac. W=Fd
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What’s the difference? • Force is the agent of change • Energy is a measure of change • Work is a way of transferring energy from one system to another
What is work? • Work= force*displac. • W=Fd • Only work if there is motion- if you push against a brick wall and it doesn’t move, you might be tired but you have done no work • Unit=Joule (unit of energy)
Work and Force at an angle • Force and work must be in same direction so if force is at angle to displacement, must use component of force parallel to displacement • So W=Fdcosθ where θ is angle between F and d
Are they work? • No- no displacement • Yes- force=g and displacement=fall • No-why? • Yes- force from engines • Teacher pushes wall and becomes exhausted • Book falls off table to floor • Waiter carries large tray across restaurant at constant v • Starship Enterprise accelerates through space So what’s with the waiter???????
What’s with the waiter? • There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). • Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement. • There was a small force when the waiter accelerated to start moving but no net force to keep it moving
+ and - work • If a force helps the motion, W is + • cosθ is + • If force opposes motion, W is - • cosθ is - • If force is perpendicular to motion, W=0 • cosθ is 0 Is work done by gravity + or -? Is work sone by person + or -?
Work = 0 • Work = 0 if: • No force • No displacement • cosθ=0 or force is perpendicular to displacement
Example… • A box slides down an inclined plane (angle = 37 degrees). The mass of the block is 35kg and the length of the ramp is 8m. • What is work done by gravity? • BE CAREFUL- angle given is NOT the angle between F and d!!!
Solution • Wgavity=(mgsinθ)d • =(35kg)(10N/kg)(sin37)(8m) • =1680J • Note work done by gravity is + b/c it is helping with motion • Also- be careful with the angle θ- usually we measure angle between F and D but in this case the angle given was the incline angle
Kinetic Energy and Work-Energy Theorem • When you do net work on an object, the result is a change in kinetic energy • K=1/2mv2 • Wtotal=ΔK
Example… • A tennis ball of mass 0.06kg is hit straight up at an initial speed of 50m/s. How high would it go? • You can solve this using the big 5 or using the work-energy theorem using gravity as the force doing negative work
Solution using WORK • Force= gravity=negative since opposes motion=mg • F and d are in opposite directions so θ = 180 so: • W=Fgcosθd=-Fgd • At the moment the ball reaches the highest point, v=0 so K=0 • Wtotal=ΔK and K=1/2mv2 • W=K0-K1=0-1/2 mv2=0-1/2(0.06kg)(50m/s)2=75J • W=-Fgd d=-W/Fg d=-W/mg • d=75J/(0.06kg)(10m/s2)=125m
Work-Energy Problems • Often an easier way to attack kinematics problems when force is involved, especially if object is at rest at start or finish since then kinetic energy=0 • Can solve for any piece of the puzzle- K, W, d, v
Potential Energy=U Energy an object has due to its position or configuration- stored energy that can be retrieved. Ex- height on a wave gives U, pulling back the string on a bow gives it U.
Gravitational Potential Energy: Ug • Potential energy due to position relative to surface of the earth • Ug=mgh • Unit = Joule
Gravitational Potential Energy: Ugand Work done by gravity • Gravity can do + or - work depending on motion • Path independent- depends on height, not path taken • Wg=mgΔh • Where h is height above arbitrary 0 pt
Examples • Physicsman (mass=60kg) scales a 40m tall rock face. What is his potential energy (relative to the ground)? • Ug=mgh=(60kg)(10N/kg)(40m)=24000J
Take it 1 step further… • If physicsman were to jump of the cliff, what would his velocity be when he hits the ground? Think…U is transformed to K • U…K…1/2mv2 • 24000J=1/2(60kg)v2 • V=28m/s
Work by Conservative vs. Nonconservative Forces • Conservative forces are path independent • Ex: gravity • Nonconservative forces depend on path • Ex: kinetic friction- longer path means more work If path is a closed loop, conservative forces W=0 but nonconservative forces do work.
Ex. of conservative vs. nonconservative • In a roller coaster doing a closed loop, the work done by gravity (conservative) is - on way up and + on way down and cancels out by the time it returns to start. • Work done by kinetic friction (nonconservative) is - (opposing motion) the entire trip so doesn’t cancel out when returns to start
Work and Energy • E=K+U • E=1/2mv2+mgh • Object’s mechanical energy is sum of kinetic and potential energies • Since U is relative to position, so is E • Wnc=ΔK+ΔU • Work done by nonconservative forces is sum of changes in K and U
Conservation of Energy • Since E=K+U, if no nonconservative forces (friction for example) act on a system then mechanical energy is conserved • Ei=Ef or Ki+Ui=Kf+Uf • If nonconervative forces act, then E is cot conserved- energy transformed into other (nonmechanical forms) • In this case Ei+Wnc=Ef • Think of what happens to your hands if you rub them together producing friction
Ex: conservation of energy • A ball of mass 2kg is gently pushed off the lab table, 5.0m above the floor. Find the speed of the ball as it strikes the floor • Ei=Ef or Ki+Ui=Kf+Uf • 0+mgh=1/2mv2+0 • v=sqrt(2gh)=10m/s
Now try one with angles :) • A skier starts from rest at the top of a 20° incline and skis a straight line to the bottom of the slope 400m down the slope. If the coefficient of kinetic friction between the skis and the snow is 0.2, calculate the skiers speed at the bottom of the run. • Hint- draw a picture and use conservation of mechanical energy- there is work done by a nonconservative force!
Skier solved Fk=μkFN=μkmgcosθ Wnc=Wfriction=Fd= -(μkmgcosθ)d Ei+Wnc=Ef 0+mgh+ -(μkmgcosθ)d=1/2mv2+0 h=dsinθ mg(dsinθ) -(μkmgcosθ)d=1/2mv2 gd(sinθ-μkcosθ)=1/2v2 v=sqrt[2gd(sinθ-μkcosθ)] =35m/s
Power • Power= rate at which work gets done= work over time • P=W/t • Since W=Fd then P=Fd/t and d/t=v • P=Fv • Unit= J/s=watt (W) • Careful not to confuse unit W (watt) with concept W (work)
Ex: Power • A mover pushes a large crate mass=75kg across the truck bed for a total distance of 6m. He exerts a steady force of 300N for 20s. What is his power output? • P=W/t P=Fd/t=(300N)(6m)/20s=90W
Ex 2: Power • What is the power output of an elevator motor that can lift 1000kg at a constant speed of 9.0m/s? Hint- what is the force? • F=mg since acting against weight • P=W/t=Fd/t=Fv • P=mgv=(1000kg)(10N/kg)(8.0m/s) • =80000W=80kW