Work Energy Power Whats the difference Force is

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 • 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? • 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 • No- no displacement • Yes- force=g and displacement=fall • No-why? • Yes- force from engines 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 35 kg and the length of the ramp is 8 m. – What is work done by gravity? – BE CAREFUL- angle given is NOT the angle between F and d!!!

Solution • • Wgavity=(mgsinθ )d =(35 kg)(10 N/kg)(sin 37)(8 m) =1680 J 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=1/2 mv 2 • Wtotal=ΔK

Example… • A tennis ball of mass 0. 06 kg is hit straight up at an initial speed of 50 m/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=Fgcosθ d=-Fgd At the moment the ball reaches the highest point, v=0 so K=0 Wtotal=ΔK and K=1/2 mv 2 W=K 0 -K 1=0 -1/2 mv 2=0 -1/2(0. 06 kg)(50 m/s)2=75 J W=-Fgd d=-W/Fg d=-W/mg d=75 J/(0. 06 kg)(10 m/s 2)=125 m

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 configurationstored 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: Ug and Work done by gravity • Gravity can do + or work depending on motion • Path independentdepends on height, not path taken • Wg=mgΔh • Where h is height above arbitrary 0 pt

Examples • Physicsman (mass=60 kg) scales a 40 m tall rock face. What is his potential energy (relative to the ground)? • Ug=mgh=(60 kg)(10 N/kg)(40 m)=24000 J

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… 1/2 mv 2 • 24000 J=1/2(60 kg)v 2 • V=28 m/s

Work by Conservative vs. Nonconservative Forces • Conservative forces are path independent – Ex: gravity • Nonconservative forces depend on path – Ex: kinetic frictionlonger 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=1/2 mv 2+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 conservedenergy 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 2 kg is gently pushed off the lab table, 5. 0 m above the floor. Find the speed of the ball as it strikes the floor • Ei=Ef or Ki+Ui=Kf+Uf • 0+mgh=1/2 mv 2+0 • v=sqrt(2 gh)=10 m/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 400 m 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=μ k. FN=μ kmgcosθ Wnc=Wfriction=Fd= -(μ kmgcosθ )d Ei+Wnc=Ef 0+mgh+ -(μ kmgcosθ )d=1/2 mv 2+0 h=dsinθ mg(dsinθ ) -(μ kmgcosθ )d=1/2 mv 2 gd(sinθ -μ kcosθ )=1/2 v 2 v=sqrt[2 gd(sinθ -μ kcosθ )] =35 m/s

• 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) Power

Ex: Power • A mover pushes a large crate mass=75 kg across the truck bed for a total distance of 6 m. He exerts a steady force of 300 N for 20 s. What is his power output? • P=W/t P=Fd/t=(300 N)(6 m)/20 s=90 W

Ex 2: Power • What is the power output of an elevator motor that can lift 1000 kg at a constant speed of 9. 0 m/s? Hint- what is the force? • F=mg since acting against weight • P=W/t=Fd/t=Fv • P=mgv=(1000 kg)(10 N/kg)(8. 0 m/s) • =80000 W=80 k. W