Summary Lecture 6 6.4Drag force Terminal velocity Work and Kinetic energy 7.7Power 8.1Potential energy 8.2/3Conservative Forces and Potential energy 8.4Conservation of Mech. Energy 8.5Potential-energy curves 8.7Conservation of Energy 6.4Drag force Terminal velocity Work and Kinetic energy 7.7Power 8.1Potential energy 8.2/3Conservative Forces and Potential energy 8.4Conservation of Mech. Energy 8.5Potential-energy curves 8.7Conservation of Energy Problems:Chap 6: 32, 33, Chap. 7: 2, 14, 50, 29, 31, Chap. 8 5, 8, 22, 29, 36, 71, 51

VISCOUS DRAG FORCE DRAG

VISCOUS DRAG FORCE Assumptions low viscosity (like air) turbulent flow What is it? like fluid friction a force opposing motion as fluid flows past object

Fluid of density  V mV m Volume hitting object in 1 sec. =AV Mass hitting object in 1 sec. =  AV momentum (p) transferred to object in 1 sec. = (  AV)V Force on object = c onst  AV 2 Area A In 1 sec a length of V metres hits the object

Fluid of density  V mV m Force on object = c onst  AV 2 Area A V Air hits object = object moves through air

V mg D D V V=0  F = mg - D  F = mg -1/2C  Av 2 D increases as v 2 until  F=0 i.e. mg= 1/2C  Av 2

0mgAv1/2C dt dv m 2   F = mg –D D mg ma = mg -D D- mg dt dv m 

When entertainment defies reality

D= ½ C  Av 2 Assume C = 1 v = 700 km h -1 Calculate: Drag force on presidents wife Compare with weight force Could they slide down the wire?

D= ½ C  Av 2 Assume C = 1 v = 700 km h -1 Calculate: The angle of the cable relative to horizontal. Compare this with the angle in the film (~30 o ) 

In working out this problem you will prove the expression for the viscous drag force

Time s Height m Real projectile motion! Throw a stone up with vel v, what is height as function of time? Drag force proportional to the square of the velocity for the ascent, mg and drag force in same direction, for the descent they are opposite. VCE Physics Real Physics

WORK You know that if I move a body through a displacement d by applying a constant force F w = Fd What if F is NOT in the direction of d? Work is energy transferred to or from an object by a force acting on the object. Energy transferred TO the object is positive work, and energy transferred FROM the object is negative work. F d What if the force is NOT constant? BUT!

If the Force is not in the direction of displacement F

w = F. d (Scalar product) F = iF x +jF y Remember for a scalar product i.i = 1 j.j = 1 i.j=0 j.i=0 w = F x d x F y d y here: d x = dd y = 0 W = Fcos  d +0 F y =Fsin  F x =Fcos  d = id x + jd y Thus w = (iF x +jF y ). (id x + jd y ) vectorsscalar y x  F d =i.iF x d x + i.jF x d y + j.iF y d x + j.jF y d y

 F 0d w = F. d =Fcos  | d | If  = 0 w = Fd if  = 90 w = 0 component of F parallel to d, multiplied by magnitude of d Work is a SCALAR: the product of 2 vectors The unit of work is JOULE

xfxf xixi What if the force is NOT constant? i.e F depends on x: F(x) Move a distance  x  w = F(x).  x In the limit as  x   0 xx F(x) F x How much work is done by F in moving object from x i to x f ? or the area under the F-x curve

F(x) x F rest = -kx Work done BY the spring The Spring Force xfxf x +ve Work = area of this triangle! F rest

Power Power is the rate of doing work If we do work  w in a time  t = F cos  |v | F cos   v F Power is a scalar: the product of F and v Unit of power is J s -1 = watt 1kw = 1000 w1 HP = 746 w F.vF.v

Kinetic Energy Work-Kinetic Energy Theorem Change in KE  work done by all forces  K   w

FF x xixi xfxf = 1/2mv f 2 – 1/2mv i 2 = K f - K i  K =  K Work done by net force = change in KE Work-Kinetic Energy Theorem vector sum of all forces

FF x xixi xfxf = 1/2mv f 2 – 1/2mv i 2 = K f - K i  K =  K Work done by net force = change in KE Work-Kinetic Energy Theorem vector sum of all forces

mg F h Lift mass m with constant velocity Work done by me (take down as +ve) = F.(-h) = -mg(-h) = mgh Work done by gravity = mg.(-h) = -mgh ________ Total work by ALL forces (  W) = 0 What happens if I let go? =K=K Gravitation and work

Compressing a spring Compress a spring by an amount x Work done by me  Fdx =  kxdx = 1/2kx 2 Work done by spring  -kxdx =-1/2kx 2 Total work done (  W) = 0 =K=K What happens if I let go? x F-kx

F f d Work done by me = F.d Work done by friction = -f.d = -F.d Total work done = 0 What happens if I let go?NOTHING!! Gravity and spring forces are Conservative Friction is NOT!! Moving a block against friction at constant velocity

Conservative Forces A force is conservative if the work it does on a particle that moves through a round trip is zero; otherwise the force is non-conservative work done for round trip: On way up: work done by gravity = -mgh On way down: work done by gravity = mgh Total work done = 0 Sometimes written as h -g Consider throwing a mass up a height h

Conservative Forces A force is conservative if the work done by it on a particle that moves between two points is the same for all paths connecting these points: otherwise the force is non-conservative. -g Each step height=  h = -mg(  h 1 +  h 2 +  h 3 +……) = -mgh Same as direct path (-mgh) Work done by gravity w = -mg  h 1 + -mg  h 2 +-mg  h 3 +… h