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## Energy

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**Energy**• Energy- The property of an object that allows it to produce change in the environment or itself. • Types of energy • Thermal • Chemical • Solar • Nuclear • Geothermal • Hydroelectric • Wind**Energy**• Kinetic Energy- The energy of an object resulting from motion. • K = 1/2 mv2 • Work - The process of changing the energy of a system • W = Fd Work is a scalar quantity • Work-Energy Theorem ∆K = W • A unit of energy/work = N•m = Joule (J)**Energy**How much work is done in lifting an object with a mass of 20 kg to a height of 3 meters? m = 20 kg d = 3 m F = mg = (20 kg)(9.8 m/s2) = 196 N W = Fd = (196 N)(3 m) = 588 N•m = 588 J**Energy**How much work is done in pulling a pail up a well by turning a crank 40 times with a radius of 80 cm with a force of 50 N? F = 50 N r = .8 m n = 40 W = (40)(50 N)(2π)(.8 m) = 10048 J W = Fd**Energy**• Power- The rate of doing work • P=W/t J/sec = watt A box weighing 200 N is lifted a distance of 12 m in 10 seconds. Find the work and power. W = F d P = W/t P = 2400 J / 10 sec = 240 W W = (200 N)(12 m) = 2400 J**70 N**30º Energy A force of 70 N is applied to a lawnmower at an angle of 30°. The lawnmower is moved 50 meters. How much work is done? F = 70 N d = 50 m W = Fd = (F cos )d W = (70 N) (cos 30°)(50 m) W = 3031 J**Energy**• Simple Machines- A device that changes the magnitude or the direction of a force needed to do the work Pulley Lever Wheel and Axle Inclined Plane Screw Wedge • Fr Resistance Force • Fe Effort Force • dr Resistance distance • de Effort Distance**Fe**Fr Lever**Fe**Fr**1/2 Fr**Fe 1/2 Fr**Fr**Fe**Fe**Fr**Fe**Fr**Energy**• Mechanical Advantage (MA) - The ratio of the resistance force to the effort force. • MA = Fr/Fe • Ideal Mechanical Advantage (IMA) - The ratio of the effort distance to the resistance distance. • IMA =de/dr**Energy**• Efficiency - The ratio of the work output to the work input • Efficiency (%) = WO/Wi x 100 = MA / IMA x 100 • An efficient machine has an MA almost equal to its IMA. No machine has an efficiency greater than 100%**Energy**A team of movers uses an inclined plane 3 meters long to move a piano weighing 3000 newtons over a doorstop .27 m high by applying a force of 500 newtons. Calculate the MA, IMA, and the efficiency. Fr = 3000 N Fe = 500 N dr = .27 m de = 3 m MA = Fr/Fe = 3000 N/ 500 N = 6 IMA = de/dr = 3 m/.27 m = 11.1 Eff = MA/IMA = 6 / 11.1 = 54% or Eff = WO/Wi = (3000 N)(.27m)/(500 N)(3m) = 54%**Energy**A force of 225 N is displaced 7.35 m which moves a weight of 1340 N to a height of .975 m Calculate the MA, IMA, and the efficiency. Fr = 1340 N Fe = 225 N dr = .975 m de = 7.35 m MA = Fr/Fe = 1340 N/ 225 N = 5.95 IMA = de/dr = 7.35m/.975m = 7.54 Eff = MA/IMA = 5.95 / 7.54 = 79% or Eff = WO/Wi = (1340 N)(.975m)/(225 N)(7.35m) = 79%**Energy**The efficiency of a lift is 72.5% What work is needed if a mass of 200 kg is displaced 5.65 m?. m = 200 kg Fr = 1960 N dr = 5.65 m Eff = WO /Wi Wi = WO / Eff Wi = (1960 N)(5.65 m)/.725 Wi = 15274 J**Energy**A resistance of 475 N is displaced .623 m by an effort force of 178 N. The efficiency of the machine is 85%. What is the displacement of the effort force? Fr = 475 N Fe = 178 N dr = .623 m Eff = .85 Eff = WO /Wi = (Fr)(dr)/(Fe)(de) de = (Fr)(dr)/(Fe)(Eff) de = (475 N)(.623 m)/(178 N)(.85) de = 1.96 m**Observations About Seesaws**• A balanced seesaw can remain horizontal • A balanced seesaw rocks back and forth easily • Two equal-weight children balance a seesaw • Two unequal-weight children don’t balance • But moving the heavy child inward helps**Physics Concept**• Rotational Inertia • A body at rest tends to remain at rest. • A body that’s rotating tends to continue rotating.**A note on positive/negative work**• If force is parallel to displacement work is positive • If force is antiparallel to displacement work is negative • If force is perpendicular to displacement work is zero.**Physical Quantities for rotational motion**• Angular Position – an object’s orientation (angle with respect to reference, i.e. horizontal • Angular Velocity – its change in angular position with time • Torque – a twist or spin (more later)**Angular rotation or**Angular position • SI unit of angular rotation is the radian • One radian is 180°/p = 57.3° • Rotation requires an axis of rotation**Angular velocity w**• SI unit: radians per second or just 1/sec • An other unit: rotations per minute (not an SI unit). • Measure of how fast an object spins • Angular velocity is a vector! • Use right hand rule to determine direction of vector • Align right thumb with axis • Align fingers with rotational movement • Thumb points into direction of angular velocity vector**Angular velocity is a vector**Right-hand rule for determining the direction of this vector. Every particle (of a rigid object): • rotates through the same angle, • has the same angular velocity, • has the same angular acceleration.**Newton’s First Lawof Rotational Motion**A rigid object that’s not wobbling and that is free of outside torques rotates at a constant angular velocity.**Center of mass**When an object is rotating freely (no fixed axis), it rotates about its center of mass**Center of Mass**• The point about which an object’s mass balances • A free object rotates about its center of mass while its center of mass follows the path of a falling object**How do we start something spinning???**We have to apply a torque to it • We need a • pivot point • lever arm • applied force • Torque = force x lever arm Lever arm is perpendicular to applied force**Torque is a vector**It has a direction and a magnitude • Use the right hand rule to figure out the direction of the torque • Thumb is torque, t • Index finger is lever, r • Middle finger is Force, F**Moment of inertia**• The moment of inertia is a measure of an object’s rotational inertia, its resistance to change in angular velocity • Analogous to mass (translational inertia) • Moment of inertia depends on • mass of object • andmass distribution (where the mass sits with respect to axis) • the axis about which the axis rotates**Physical Quantities**• Angular Position – an object’s orientation • Angular Velocity – its change in angular position with time • Torque – a twist or spin • Angular Acceleration – its change in angular velocity with time • Moment of Inertia – measure of its rotational inertia**Newton’s Second Lawof Rotational Motion**The torque exerted on an object is equal to the product of that object’s moment of inertia times its angular acceleration. The angular acceleration is in the same direction as the torque. Torque = Moment of Inertia · Angular Acceleration**Physics Concept**• Net Torque • The sum of all torques on an object. • Determines that object’s angular acceleration.**Torques are equal (40 kg)(1 m) = x (1 m)**The mass of the plank is 40 kg. Beth has no Friends, she Has to teeter totter by herself 1 m**Summary: Angular and linear quantities**Linear motion Rotational motion Angular position (angle):q Position:x Velocity:v Angular velocity:w Angular Acceleration:a Acceleration:a Torque: Force: Mass:m Moment of inertia:I Momentum:mv Angular Momentum:mvr**Law of Conservation of Angular Momentum**• In a closed system, angular momentum is conserved • (mvr)before = (mvr)after**Energy**Kinetic Energy – Energy of an object due to motion K = ½ mv2 Potential Energy – Energy of an object due to position • Gravity • Spring • Battery • Chemical Bonds Anything that can be converted to energy Ug= mgh**Energy**A car with a mass of 1000 kg moves with a velocity of 20 m/s. What is its kinetic energy? m = 1000 kg v = 20 m/s K = ½ mv2 K = ½ (1000 kg)(20 m/s)2 = 200000 kg m2/s2 = 200000 J**A**33 m C 15 m B 10 m Energy A roller coaster of mass 3000 kg goes from point A to point B to point C. a. Find the potential energy at A. b. Find the potential energy at C.**A**33 m C 15 m B 10 m Energy UA = mgh = (3000 kg)(9.8 m/s2)(33 m) = 970,200 J UC = mgh = (3000 kg)(9.8 m/s2)(15 m) = 441,000 J**10 N**20 m 10 N 10 m 10 N Energy KE = 0 J PE = 200 J Law of Conservation of Energy – The total amount of energy in a closed system is constant. Mechanical Energy E = K + Ug KE = 100 J PE = 100 J Conservation of Mechanical Energy Kbefore+Ug before = Kafter+Ug after KE = 200 J PE = 0 J**Energy**A ball with a mass of 20 kg falls from a roof 10 meters high. • What is the potential energy? • What is the kinetic energy of the ball when it reaches the ground? • What is the speed of the ball when it hits the ground? a . Ug = mgh = (20 kg)(9.8 m/s2)(10 m) = 1960 J b. Kbottom= Ug top Kbottom = 1960 J c. K = ½ mv2**Energy**Elastic Collision Collision in which the total kinetic energy of two objects is the same after the collision as before. Inelastic Collision Collision in which some of the kinetic energy is changed to another form of energy.