Physics 111 Lecture 6 Work-Energy -Power Dr.Ali ÖVGÜN EMU Physics Department www.aovgun.com
Why Energy? q Why do we need a concept o energy? q The energy approach to describing motion is particularly useul when Newton s Laws are diicult or impossible to use. q Energy is a scalar quantity. It does not have a direction associated with it. October 7-13, 013
Kinetic Energy q Kinetic Energy is energy associated with the state o motion o an object q For an object moving with a speed o v K = 1 mv q SI unit: joule (J) 1 joule = 1 J = 1 kg m /s October 7-13, 013
Kinetic Energy or Various Objects KE = 1 mv October 7-13, 013
K = 1 mv Why? October 7-13, 013
Work W q Start with 1 mv 1 mv 0 = FxΔx Work W q Work provides a link between orce and energy q Work done on an object is transerred to/rom it q I W > 0, energy added: transerred to the object q I W < 0, energy taken away: transerred rom the object October 7-13, 013
Deinition o Work W q The work, W, done by a constant orce on an object is deined as the product o the component o the orce along the direction o displacement and the magnitude o the displacement W ( F cosθ )Δ x n F is the magnitude o the orce n Δ x is the magnitude o the object s displacement n θ is the angle between =!! F Δ x!! F and Δx October 7-13, 013
Work Unit q This gives no inormation about n the time it took or the displacement to occur n the velocity or acceleration o the object q Work is a scalar quantity q SI Unit n Newton meter = Joule n N m = J mv n J = kg m / s = ( kg m / s ) m W ( F cosθ )Δ x = 1 1 mv0 = ( F cosθ ) Δx!! F Δ x October 7-13, 013
Work: + or -? q Work can be positive, negative, or zero. The sign o the work depends on the direction o the orce relative to the displacement!! W ( F cosφ) s = F s q Work positive: i 90 > φ> 0 q Work negative: i 180 > φ> 90 q Work zero: W = 0 i φ= 90 q Work maximum i φ= 0 q Work minimum i φ = 180 October 7-13, 013
Work done by a Gravitational Force q Gravitational Force n Magnitude: mg n Direction: downwards to the Earth s center q Work done by Gravitational Force r r W = FΔ rcosθ = F Δ W net = 1 mv 1 mv 0 W g = mgδr cosθ November 3, 008
Example: When Work is Zero q A man carries a bucket o water horizontally at constant velocity. q The orce does no work on the bucket q Displacement is horizontal q Force is vertical q cos 90 = 0 W ( F cosθ ) Δx October 7-13, 013
Example: Work Can Be Positive or Negative q Work is positive when liting the box q Work would be negative i lowering the box n The orce would still be upward, but the displacement would be downward October 7-13, 013
Work Done by a Constant q Force The work W done a system by an agent exerting a constant orce on the system is the product o the magnitude F o the orce, the magnitude Δr o the displacement o the point o application o the orce, and cosθ, where θ is the angle between the orce and displacement vectors: W!! F Δr = FΔr cosθ W I W III I III F! = 0 Δr! F! Δr! = FΔr F! Δr! II W II W IV = FΔr F! Δr! IV = FΔr cosθ October 7-13, 013
Work and Force q An Eskimo returning pulls a sled as shown. The total mass o the sled is 50.0 kg, and he exerts a orce o 1.0 10 N on the sled by pulling on the rope. How much work does he do on the sled i θ = 30 and he pulls the sled 5.0 m? W = ( F cosθ ) Δx = (1.0 10 = 5. 10 J N)(cos30! )(5.0m) October 7-13, 013
Work Done by Multiple Forces q I more than one orce acts on an object, then the total work is equal to the algebraic sum o the work done by the individual orces W net = W by individual orces n Remember work is a scalar, so this is the algebraic sum W net = W + W + W = ( F cosθ ) Δr g N F October 7-13, 013
Problem Solving Strategy q Identiy the initial and inal positions o the body, and draw a ree body diagram showing and labeling all the orces acting on the body q Choose a coordinate system q List the unknown and known quantities, and decide which unknowns are your target variables q Calculate the work done by each orce. Be sure to check signs. Add the amounts o work done by each orce to ind the net (total) work W net q Check whether your answer makes sense October 7-13, 013
Kinetic Energy q Kinetic energy associated with the motion o an object 1 K = mv q Scalar quantity with the same unit as work q Work is related to kinetic energy 1 mv 1 mv 0 = F netδ x W = KE KE = ΔKE net i October 7-13, 013
Work-Energy Theorem q When work is done by a net orce on an object and the only change in the object is its speed, the work done is equal to the change in the object s kinetic energy W net = K K i = ΔK n Speed will increase i work is positive n Speed will decrease i work is negative W net = 1 mv 1 mv 0 October 7-13, 013
Potential Energy q Potential energy is associated with the position o the object q Gravitational Potential Energy is the energy associated with the relative position o an object in space near the Earth s surace q The gravitational potential energy n n n n PE mgy m is the mass o an object g is the acceleration o gravity y is the vertical position o the mass relative the surace o the Earth SI unit: joule (J) November 3, 008
Reerence Levels q A location where the gravitational potential energy is zero must be chosen or each problem n The choice is arbitrary since the change in the potential energy is the important quantity n Choose a convenient location or the zero reerence height n oten the Earth s surace n may be some other point suggested by the problem n Once the position is chosen, it must remain ixed or the entire problem November 3, 008
q PE = mgy q W g = mg( y Work and Gravitational = FΔy cosθ = q Units o Potential Energy are the same as those o Work and Kinetic Energy W grav ity Potential Energy y i = ) PE mg( y i i PE y )cos0 November 3, 008
xtended Work-Energy Theorem q The work-energy theorem can be extended to include potential energy: Wnet = KE KEi = ΔKE W grav ity PE q I we only have gravitational orce, then KE KE i = = q The sum o the kinetic energy and the gravitational potential energy remains constant at all time and hence is a conserved quantity PE i i KE + PE = PE + i PE PE KE i W net = W gravity November 3, 008
Extended Work-Energy Theorem q We denote the total mechanical energy by q Since E = KE + PE KE + PE = PE + i KE i q The total mechanical energy is conserved and remains the same at all times 1 1 mv i + mgyi = mv + mgy q I there is riction orce PE + KE = E + KE + PE E i i Lostbyriction Lostbyriction = d November 3, 008
Problem-Solving Strategy q Deine the system q Select the location o zero gravitational potential energy n Do not change this location while solving the problem q Identiy two points the object o interest moves between n One point should be where inormation is given n The other point should be where you want to ind out something November 3, 008
Platorm Diver q A diver o mass m drops rom a board 10.0 m above the water s surace. Neglect air resistance. q (a) Find is speed 5.0 m above the water surace q (b) Find his speed as he hits the water November 3, 008
Platorm Diver q (a) Find is speed 5.0 m above the water surace 1 1 mv i + mgyi = mv + mgy 1 0 + gy i = v + mgy v = g( y i y ) = (9.8m / s )(10m 5m) = 9.9m / s q (b) Find his speed as he hits the water 1 0 + mgy i = mv + 0 v = gyi =14m / s November 3, 008
Power q Work does not depend on time interval q The rate at which energy is transerred is important in the design and use o practical device q The time rate o energy transer is called power q The average power is given by W P = Δt n when the method o energy transer is work October 7-13, 013
Units o Power qthe SI unit o power is called the watt n 1 watt = 1 joule / second = 1 kg. m / s 3 qa unit o power in the US Customary system is horsepower n 1 hp = 550 t. lb/s = 746 W qunits o power can also be used to express units o work or energy n 1 kwh = (1000 W)(3600 s) = 3.6 x10 6 J October 7-13, 013
Problems October 7-13, 013
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