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Summary Lecture 6 6.4 7.1-7.6 Drag force Terminal velocity Work and Kinetic energy 7.7 Power 8.1 Potential energy 8.2/3 Conservative Forces and Potential energy 8.4 Conservation of Mech. Energy 8.5 Potential-energy curves 8.7 Conservation of Energy Problems:Chap 6: 32, 33, Chap. 7: 2, 14, 50, 29, 31, Chap. 8 5, 8, 22, 29, 36, 71, 51 http://webraft.its.unimelb.edu.au/640141/pub/lectures/mechanics/lecture6.pdf VISCOUS DRAG FORCE VISCOUS DRAG FORCE What is it? like fluid friction a force opposing motion as fluid flows past object Assumptions low viscosity (like air) turbulent flow Vm Area A Fluid of density In 1 sec a length of V metres hits the object 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 = const AV2 p F t Vm Area A Fluid of density Air hits object = object moves through air V Force on object = const AV2 SF = mg - D SF = mg -1/2CAv2 V=0 mg D V D increases as v2 until SF=0 mg i.e. mg= 1/2CAv2 D V mg 2mg v te rm CA 2 v te rm 2mg CA D SF = mg –D ma = mg -D mg dv mg - D m dt dv m + 1/2CAv 2 - mg 0 dt v [ 2gm (1 - e cA - 2m t cA )]1 / 2 2gm 1 / 2 v [ ] cA v [ 2gm (1 - e cA - 2m t cA )]1 / 2 When entertainment defies reality Calculate: D= ½ CAv2 Assume C = 1 v = 700 km h-1 Drag force on presidents wife Compare with weight force Could they slide down the wire? Calculate: D= ½ CAv2 Assume C = 1 v = 700 km h-1 The angle of the cable relative to horizontal. Compare this with the angle in the film (~30o) In working out this problem you will prove the expression for the viscous drag force 1 F CAv 2 2 Real projectile motion! Real Physics VCE Physics Height m 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. Time s http://www.colorado.edu/physics/phet/web-pages/simulations-base.html WORK 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. You know that if I move a body through a displacement d by applying a constant force F w = Fd F d BUT! BUT! What if F is NOT in the direction of d? What if the force is NOT constant? If the Force is not in the direction of displacement y scalar F vectors w=F.d F = iFx +jFy d = idx + jdy Fy=Fsin x Fx=Fcos (Scalar product) d Thus w = (iFx +jFy) . (idx + jdy) =i.iFxdx + i.jFxdy + j.iFydx + j.jFydy Remember for a scalar product i.i = 1 j.j = 1 i.j=0 j.i=0 w = Fxdx here: +0 +0 + Fydy dx = d W = Fcos d dy= 0 +0 w=F.d F =Fcos |d | 0 d component of F parallel to d, multiplied by magnitude of d If = 0 w = Fd if = 90 w = 0 Work is a SCALAR: the product of 2 vectors The unit of work is JOULE What if the force is NOT constant? i.e F depends on x: F(x) How much work is done by F in moving object from xi to xf? Move a distance x F(x) F w = F(x). x xf w F ( x ).x xi x In the limit as x 0 xi xf w F( x ).dx xi xf or the area under the F-x curve x The Spring Force +ve F(x) Frest x xf Work done BY the spring x Work = area of this triangle! xf w F ( x ) dx xi xf w - kx dx xi x f w - k x dx - k 1/ 2 x 2 xi 1 2 if xi 0 then w - kx f 2 Work done BY the spring xf xi Power Power is the rate of doing work If we do work w in a time t Δw pav Δt pinst limit Δt 0 Δw dw Δt dt Δw F. Δx p inst Δt Δt F F cos F.v = F cos |v | Power is a scalar: the product of F and v Unit of power is J s-1= watt 1kw = 1000 w 1 HP = 746 w v Kinetic Energy Work-Kinetic Energy Theorem Change in KE work done by all forces K w Work-Kinetic Energy Theorem w x F .dx xf i x ma.dx xf i x x i f dv m .dx dt x x i f dx m .dv dt vector sum of all forces SF xi xf m v .dv m[1 / 2v ] 2 vf vi vf vi = 1/2mvf2 – 1/2mvi2 = Kf - Ki = K Work done by net force = change in KE x Work-Kinetic Energy Theorem w x F .dx xf i x ma.dx xf i x x i f dv m .dx dt x x i f dx m .dv dt vector sum of all forces SF xi xf m v .dv m[1 / 2v ] 2 vf vi vf vi = 1/2mvf2 – 1/2mvi2 = Kf - Ki = K Work done by net force = change in KE x Gravitation and work Work done by me (take down as +ve) h F mg Lift mass m with constant velocity = F.(-h) = -mg(-h) = mgh Work done by gravity = mg.(-h) = -mgh ________ Total work by ALL forces (W) = 0 =K What happens if I let go? Compressing a spring Compress a spring by an amount x F -kx x Work done by me Fdx = kxdx = 1/2kx2 Work done by spring -kxdx =-1/2kx2 Total work done (W) = 0 =K What happens if I let go? Moving a block against friction at constant velocity f F d Work done by me = F.d Work done by friction = -f.d = -F.d Total work done What happens if I let go? = 0 NOTHING!! Gravity and spring forces are Conservative Friction is NOT!! 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 Consider throwing a mass up a height h -g h work done for round trip: On way up: work done by gravity = -mgh On way down: work done by gravity = mgh Total work done Sometimes written as F.ds 0 = 0 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. Work done by gravity Each step height=h w = -mgh1+ -mgh2+-mgh3+… = -mg(h1+h2+h3 +……) -g h = -mgh Same as direct path (-mgh)