Unit2_Energy_Gravity_ProjectileMotion_wk4_fixed0521.pptx

Energy  The combination of energy and matter make up the universe:  Matter is substance  Energy is the mover of substance  Matter is stuff that we can smell and feel while energy is more abstract Work  Work is the component of force in the direction of motion...

Energy  The combination of energy and matter make up the universe:  Matter is substance  Energy is the mover of substance  Matter is stuff that we can smell and feel while energy is more abstract Work  Work is the component of force in the direction of motion times the distance moved W = Fd  Units of work are Newton-meters (N*m), also known as Joules (J)  Example: A weight lifter that holds weight over his head is not doing any work on the barbell – lifting the barbell is a different story. Power  Power is the rate at which energy is expended  Power = (work done) / (time interval)  Units of Power – Joules per second (J/s), also known as a watt (W).  Examples: A high-power engine does work rapidly– twice the power means that it can do twice the work in the same amount of time. Mechanical Energy  Energy due to the position of something or the movement of something  Exists in the forms of potential and kinetic energy and sums of both Potential Energy  The energy that something possesses because of its position  Gravitational potential energy = weight * height  PE = mgh  Kinetic Energy  The energy of motion  Kinetic Energy = ½ mv2  KE = ½ mv2  The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest:  Net force x distance = kinetic energy  Fd = ½ mv2 Work-Energy Theorem  Thework done on an object equals the change in kinetic energy of the object  Work = D KE  Workcan also transform other forms of energy to a system  Work can change the PE of a mechanical device, the heat energy in a thermal system, or the electrical energy in an electrical device  Conservation of Energy  Energy cannot be created or destroyed;  it may be transformed from one form into another  the total amount of energy never changes  Machines  A device, such as a lever or pulley, that increases (or decreases) a force or simply changes the direction of a force.  Lever – Simple machine consisting of a rigid rod pivoted at a fixed point called the fulcrum  Conservation of energy of machines  The work output of any machine cannot exceed the work input.  In an ideal machine, where no energy is transformed into thermal energy:  Work input = work output  (Fd)input = (Fd)output  Efficiency  The percentage of the work put into a machine that is converted into useful work output.  [(Useful energy output) / (total energy output)]*100  Comparison of KE and Momentum 1200 1000 800  Momentum = mv Momentum (kg*m/s) 600  directly proportional to v 400 200  KE = ½ mv2 0 0 10 20 30 40 50 60 70 80  proportional to v squared velocity (m/s) 45000 40000 35000 30000 25000 KE (J) 20000 15000 10000 5000 0 0 10 20 30 40 50 60 70 80 velocity (m/s) Example Problem 1  (1) The second floor of a house is 6m above street level. How much work is required to life a 300-kg refrigerator to the second-floor level?  Answer:  W = ∆E = ∆mgh = 300 kg 10 N/k 6 m = 18,000 J. Example Problem 2  (3) which produces the greater change in kinetic energy: exerting a 10-N force for a distance of 5m, or exerting a 20-N force over a distance of 2m (assuming all the work goes into KE)?  Answer:  The work done by 10 N over a distance of 5 m = 50 J. That by 20 N over 2 m = 40 J. So the 10-N force over 5 m does more work and could produce a greater change in KE. Example Problem 3  (5) A lever is used to lift a heavy load. When a 50-N force pushes one end of the lever down 1.2m, the load rises 0.2m. Calculate the weight of the load.  Answer:  (F d)in=(F d)out  50N 1.2 m = W 0.2 m  W = [(50 N)(1.2 m)]/0.2 m = 300 N. Example Problem 4  In a hydraulic machine it is observed that when the small piston is pushed down 10 cm, the large piston is raised 1 cm. If the small piston is pushed down with a force of 100 N, what is the most force that the large piston could exert?  Answer:  (Fd)input = (Fd)output  (100 N x 10 cm)input = (? x 1 cm)output  So we see that the output force is 1000 N (or less if the efficiency is less than 100%).  Example Problem 5 Consider the inelastic collision between the tow freight cars in the previous chapter (figure 6.14) see below. The momentum before and after the collision is the same. The KE, however, is less after the collision than before the collision,. How much less, and what becomes of this energy?  Answer:  The freight cars have only half the KE possessed by the single car before collision. Here’s how to figure it:  KEbefore = 1/2 m v2  KEafter = 1/2 (2m)(v/2)2 = 1/2 (2m)v2/4 = 1/4 mv2.  What becomes of this energy? Most of it goes into nature’s graveyard— thermal energy.  Example Problem 6  How many kilometers per liter will a car obtain if its engine is 25% efficient and it encounters an average retarding force of 500 N at highway speed? Assume that the energy content of gasoline is 40 MJ/liter (aka. 40,000,000 J/liter).  Answer:  At 25% efficiency, only 1/4 of the 40 megajoules in one liter, or 10 MJ, will go into work. This work is F x d = 500 N x d = 10 MJ.  Solve this for d and convert MJ to J, to get  d = 10 MJ/500 N = 10,000,000 J/500 N = 20,000 m = 20 km.  So under these conditions, the car gets 20 kilometers per liter (which is 47 miles per gallon). Gravity  Law of Universal Gravitation:  Everything pulls on everything else in a way that only involves mass and distance  F= (Gm1m2)/(d2)  Where G is the constant  G=6.67 x 10-11 N*m2/kg2 Gravity and Distance: Inverse-Square Law  F= (Gm1m2)/(d2)  We can model the inverse square relationship by looking at the thickness of paint with distance from a paint can  Weight and Weightlessness  Remember that the weight that registers on your bathroom scale is indicative of the support force  For situations were there is an external acceleration to the system containing you and the scale, your weight will appear different  Astronauts in orbit are without a support force and are in a sustained state of apparent weightlessness (they are continually in free fall) Example Problem 7  Findthe change in the force of gravity between two planets when the distance between them is decreased by a factor of five.  Answer:  FromF = GmM /d2, 15 of d squared is 1/25th d2, which means the force is 25 times greater. Projectile Motion  During projectile flight, velocities in both the vertical and horizontal directions can be considered Projectiles launched horizontally  Projectiles maintain the same horizontal velocity (vx) throughout their path (neglecting air resistance), while the vertical velocity (v y) usually changes due to gravity, etc. Projectiles launched at an angle  Different launching angles change the initial proportions of the velocity in the horizontal and vertical directions, and this results in different overall paths  Air Drag and Projectile Motion  Without the effects of air, the max range for a projectile would happen at 45 degrees, but for sports these effects are significant resulting in other optimum angles:  Baseball – less than 45  Golf – less than 38 (more effected by spin and air drag)  Shot put and Javelin (~45)  Fast-Moving Projectiles - Satellites  As mentioned previously, astronauts in orbit are continually in a state of free fall  In actuality an Earth satellite is simply a projectile that falls around the Earth rather than into it.  Example Problem 8  A ball is thrown horizontally from a cliff at a speed of 10 m/s. What is its speed one second later?  Answer  One second after being thrown, its horizontal component of velocity is 10 m/s, and its vertical component is also 10 m/s. By the Pythagorean theorem, V = √(102 + 102) = 14.1 m/s. (It is moving at a 45° angle.)  Example Problem 9  A cannonball shot with an initial velocity of 141 m/s at an angel of 45 degrees follows a parabolic path and hits a balloon at the top of its trajectory. Neglecting air resistance, how fast is the cannonball going when it hits the balloon?  Answer:  100 m/s. At the top of its trajectory, the vertical component of velocity is zero, leaving only the horizontal component. The horizontal component at the top or anywhere along the path is the same as the initial horizontal component, 100 m/s (the side of a square where the diagonal is 141).  Example Problem 10  John and Tracy look from their 80-m high-rise balcony to a swimming pool below– not exactly below, but rather 20m out from the bottom of the building. They wonder how fast they would have to jump horizontally to succeed in reaching the pool. What is the answer?  Answer:  John and Tracy’s horizontal jumping velocity will be the horizontal distance traveled divided by the time of the jump. The horizontal distance will be a minimum of 20 m, but what will be the time? Aha, the same time it would take John and Tracy to fall straight down! From Table 3.3 we see such a fall would take 4 seconds. Or we can find the time from  d = 5t2, where rearrangement gives t = √(d/5) = √(80/5) = 4 s. Example Problem 11  Calculate a person’s hang time who moves horizontally 3 m during a 1.25-m high jump. What is the hang time when moving 6 m horizontally during this jump?  Answer:  Hang time depends only on the vertical component of initial velocity and the corresponding vertical distance attained. From d = 5t2 a vertical 1.25 m drop corresponds to 0.5 s (t = √2d/g = √2(1.25)/10 = 0.5 s).  Double this (time up and time down) for a hang time of 1 s. Hang time is the same whatever the horizontal distance traveled.