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Chapter 6 Work and Energy • Work done by a constant force • Work-energy theorem, kinetic energy • Gravitational potential energy • Conservation of mechanical energy • Conservative and non-conservative forces • Work-energy theorem and non-conservative forces • Power • Work done by a variable force IMPORTANT: YOU NEED TO KNOW HOW TO APPLY THESE 1) v vo at 2) 1 2 x xo vot at 2 3) 1 x xo v vo t 2 4) 2a x xo v 2 vo2 F 2ax v vo 2 x m 1 1 2 Fx mv mv o2 2 2 2 2 Work-Energy theorem 1 Kinetic Energy : EK mv 2 2 Work done = change in kinetic energy 1 1 2 Fx mv mv o2 2 2 Fx = “work” done by the force = force times displacement The work changes the speed of the object thereby increasing its kinetic energy. Initial kinetic energy: 1 KE o mv o2 2 Final kinetic energy: 1 KE mv 2 2 Work done: W KE KEo KE The unit of work and energy: Joule (J) 1 J 1 N m Work Done by a Constant Force The mechanical work done is equal to the force times the distance something is moved: W Fs 1 N m 1 joule J Work Only the component of the force in the direction of the displacement Counts: Fx F cos a m m 1 1 F cos 2 2 2 2 So, 2 x v vo Fx cos mv mv o 2 2 m Work done: W Fx cos KE WORK-ENERGY THEOREM Work done: (force in direction of displacement) (displacement) A friction force fk = 70 N acts on the skis. Initial speed, v0 = 3.6 m/s Find final speed, vf =?, after skiing 57 m down the slope Work-Energy Theorem: W = Fnet (displacement) = KE F m g sin 25 fk net force down the slope W Fnet d KEf KEo KE F m g sin 25 fk 58 kg g sin 25 70 N 170 N Displacement d = 57 m So, W 170 N 57 m 9690 J 1 1 2 2 KE o mv o 58 3.6 375 .8 J 2 2 So,W 9690 J KEf 375.8 J KEf 10066 J 1 2 and KEf (58)vf 10066 J vf 18.6 m/s 2 Check: 2 vf vo 2 Fnet / m d 3.6 2 57 170 / 58 2 2 vf 18.6 m/s Serway, Physics for Scientists and Engineers Problem 6.10 A 55-kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 150 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force. FP 150 N FN m fk ,x 0.25FN ,y w 7m FP 150 N FN m fk ,x 0.25FN ,y w 7m No work is done in holding an object at rest: • no displacement, no work • also no net work in lifting an object up, then returning it to its starting point as the net displacement is zero Work done in lifting an object A force F lifts the mass at constant speed through a height h. The displacement is h The applied force in the direction of the displacement is: F mg (no acceleration) The work done by the force F is: W Fh mgh But the kinetic energy has not changed – the gravity force mg has done an equal amount of negative work so that the net work done on the mass is zero. Work done in lifting an object Alternative view: define a different form of energyGravitational potential energy: PE mgy Define: Mechanical energy = kinetic energy + potential energy Mechanical energy: Then: 1 E mv 2 mgy 2 Work done by applied force, F, is (change in KE) + (change in PE) So, W Fh KE PE Check, using forces and acceleration Net upward force on the mass is F mg Apply Newton’s second law to find the acceleration: F mg ma So, a F mg m From equation of kinematics number 4: v 2 vo2 2ah h vo 2F mg m 2 So, 1 1 2 mv mv o2 Fh mgh 2 2 W Fh KE PE Two ways to look at the problem 1) A net upward force (F - mg) does work and moves the mass upward and changes its kinetic energy: (F mg)h KE Work - Energy Theorem 2) An applied force F does work and changes the kinetic and potential energies of the mass: W Fh KE mgh KE PE The second is more powerful as it can be turned into a general principle that: Work done by an applied force = change in mechanical energy If no work is done, then the mechanical energy is conserved! W Fh KE PE If there is no net external force, W = 0 and so that 0 KE PE KE PE As the mass falls and loses potential energy, it gains an equal amount of kinetic energy. Potential energy is converted into kinetic energy. Energy is conserved overall. Conservation of Mechanical Energy In the absence of applied forces and friction: The work done by a net zero applied force is zero So, 0 KE PE And KE + PE = E = total energy is constant Total Energy is Conserved in a Closed System Other kinds of potential energy: • elastic (stretched spring) • electrostatic (charge moving in an electric field) No applied (i.e. external) forces E = KE + PE = constant KE = mv2/2 PE = mgy So E = mv2/2 + mgy = constant, until the ball hits the ground Check: Problem 6.40 A particle, starting from point A in the drawing, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 4.00 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A. Conservative Forces Gravitational potential energy depends only on height The difference in PE, mg(ho - hf) is independent of path taken Gravity is a “conservative force” Conservative Forces Alternative definitions of conservative forces: • The work done by a conservative force in moving an object is independent of the path taken. (Compare pushing a crate across the floor – the most direct path requires the least work. Friction is not a conservative force). • A force is conservative when it does no net work in moving an object around a closed path, ending up where it started. In either case, the potential energy due to a conservative force depends only on position. Examples of conservative forces: • Gravity, elastic spring force, electric force vo 0 Skier starts from rest A v B Problem 6.42: What must be the height h if the skier just loses contact with the snow at the crest of the second hill at B? FN v B mg r 36 m