advertisement

Conservative and Nonconservative Forces Work done by gravity on a closed path is zero: 1 Conservative and Nonconservative Forces Work done by friction on a closed path is not zero: 2 Vacuum Ball is dropped from rest at a height h and experiences only a gravitational force. Gravity does positive work transforming Potential Energy into Kinetic Energy vi = 0 mg h h/2 mg h/4 mg 3 Canola Oil 4 The oil produces a constant non-conservative force FNC against the ball opposite to the direction of motion vi = 0 mg FNC h FNC mg h/2 h/4 mg If FNC=mg, the forces balances and acceleration ceases, and the ball moves downward at constant speed Change in potential energy implies motion as conservative forces depend only on position ∆U Work Done=Energy Flow ∆KE ∆U Work Done=Energy Flow ∆KE WNC 5 While traveling on Delaware Avenue at 25 mph during a snowy day, you approach an intersection with a red light. It is known the coefficient of friction for your car tires is 0.20 in these conditions. What is the minimum distance to stop the car? Flat road d f distance to stop the car frictional force between tires and snowy road 6 Two blocks are connected by a string that passes over a pulley . The block of mass m1 lies on a horizontal surface and is connected to a spring with a force constant k. The system is released from rest when the spring is unstretched. y h If the hanging block of m2 falls a distance h before coming to rest, calculate the coefficient of kinetic friction μk between the block of m1 and the surface. Initial and final configuration is at rest therefore Two forms of potential energy: gravitation and the spring. Initial condition for the spring: unstretched at x=0 Initial PE for the spring is zero Final condition: block m2 falls a distance h and consequently the spring stretched to x=h. x KEi = KEf=0 Ui = Ugi + Usi= m2gh + 0 Uf = Ugf + Usf= 0 + ½kh2 7 N KEi = KEf=0 Ui = Ugi + Usi= m2gh + 0= m2gh Uf = Ugf + Usf= 0 + ½kh2= ½kh2 y Total initial mechanical energy x Total final mechanical energy Frictional force for the block on the flat surface is the product of the normal force m1g and the coefficient of friction μk Friction does negative work on the system Substituting for Wnc Solve for μk 8 Clicker You see a leaf falling to the ground with constant speed. When you first notice it, the leaf has initial total energy PEi + KEi. You watch the leaf until just before it hits the ground, at which point it has final total energy PEf + KEf. How do these total energies compare? a) PEi + KEi > PEf + KEf b) PEi + KEi = PEf + KEf c) PEi + KEi < PEf + KEf d) impossible to tell from the information provided As the leaf falls, air resistance exerts a force on it opposite to its direction of motion. This force does negative work, which prevents the leaf from accelerating. This frictional force is a nonconservative force, so the leaf loses mechanical energy as it falls, and its final total mechnical energy is less than its initial total energy. 9 Potential Energy Curves and Equipotentials The curve of a hill or a roller coaster is itself essentially a plot of the gravitational potential energy: The potential energy U and kinetic energy K add to the total energy E0 (dashed line) at all x values. K vanishes at A and D, which are the turning points of the motion. 11 Contour maps are a 2D form of gravitation potential energy curves: 450 350 250 150 50 Contour maps are lines of equal elevation or potential energy drawn on a map. Each line passes through a potential energy of a given value. 1) Lines may never cross or touch. 2) Energy spacing is the same for adjacent lines If an circular object is placed somewhere on the contour map at rest, it will move toward a lower potential energy line Recall the problem from chapter 6 However what can we say about the motion after the two separate.? Consider the following problem: From Newton’s 3rd law, we know the force that the bow acting exerts on the arrow is matched by a force in the opposite direction on the bow and the archer. This force causes the archer to slide backwards with the speed requested by the above problem. The acceleration, forces or bow energy are not given so we are unable to solve this problem using models we have learned so far. In an isolated system of two particles, the only force on one particle is that from the other particle. Newton’s third law action-reaction pair Newton’s third law action-reaction pair Newton’s 2nd law states there must be an acceleration associated with the forces Replace the acceleration with time rate of change of the velocity If mA and mm are constant, we can then rewrite the equation The time rate change of the sum =0, therefore it must be constant = constant What is this quantity?? Linear momentum of a particle or an object of mass m with a velocity is defined as the product of the mass and velocity Like velocity, momentum is a vector quantity, possessing a direction as well as a magnitude. We can now solve this problem Final and Initial total momentum of archer and arrow must equal Pi=Pf Initial conditions: Vmi =0 and VAi =0 Solve for the archer’s final velocity The net force acting on an object is the rate of change of its momentum: The General form of Newton’s 2nd law If the net force is zero, the total momentum does not change: Newton’s 1st Law