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Mechanical Energy More en the Work-Energy Theorem In a mechanical "system", the quantity "Mechanical Energy" is often used. It has the symbol Em. A EK - EK2 - EK1 Wnet * A EK W n e t =E K 2 - EK1 Em= E K + E g Wnet = | Ad | | Fnet | cos9 Remember Ee If a mechanical "system" loses no energy to the surroundings in the form of heat, light, sound etc over time, the total mechanical energy remains constant or is conserved. This is called the[.conservationtA mechanical energy. ., , ir*!**— "P'H O*^ -r-^—^-^- Therefore we can create a new work-energy formula... I Ad] |F n e t |cose = E K 2 - EK1 ;.-•.; + • i v v;,x'j.,.. •;••••••••<. :•>•.-. ..• v..-^.: v . ,'f,> :•:., '0 !,< .:• >;i'•;•••> ><? >•>;'•• - : , -• •yfi:'t;*k-: J>X'-, '.-:, >:{U"' : '^- . • • • Alternate Definition of Work • .s1 ^ Suppose an object is moving in a direction given by its displacement as shown. Suppose the net force is acting t as shown. What component of the net force does no work on.the object? is perpendicular to net 90° A d . It does no work on the object. Depending on 6 , net I along A d can only Ad apeed up or slow down the net I object. Therefore, work done We can write ... ..by the net force only changes vv the kinetic energy of the net -"^K object. This is called the work-energy theorem. net go|0 Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg = 0m/s v1 = 72.0 km/h = 20.0 m/s f = 5850N 9 = 180 Unknown: | Ad | = ? Formula: Sub: Ad | Fnet cosO = E K 2 - EK1 Note f K supplie the net force in this Ad | | Fnet | cose = E K2 ease. 2 | Ad | (5850) cosl 80° = 0 - (1260) (20.0) 12 - 5850 | Ad | = - 252000 I Ad I = 43.1 m Example: A 72,0 N applied force compresses a vertical spring 12,0 cm. A small 21.0 g marble is placed at rest on the compressed.spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? ' . . • Apply conservation of mechanical inical energy: Em (start)= Em (max height) tti :htt 0 K1 E g1 E e1 = EK2 + 0 + KX^/2 = KX2/2 = mgh 600(.12) 2 /2 = 0.021(10)h hmax = 20.6 m j j _ E g2> + ^e2 3h+ 0 X= 0.120m Choose a zero reference level for gravitational potential energy. Why can't it be ground? hmax =? K = 600 N/m t* Work done by an applied force in a system with friction Critical thinking: Can you explain why this equation is valid to find the work done by an applied force introduced to a system where energy is lost to the surroundings due to friction? Fs = 72 N I Wapplied = AEm + ®r 0.021 kg '^^^^^^Mi • • 1 ground Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? Find K first: K=FS/X = K = 72 70.120 K = 600 N/m Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - fk d The negative sign represents a loss of heat energy from the system to the surroundings How can we modify the conservation of mechanical energy formula to take friction into account? m= 0.021 kg 0.120m Choose a zero reference level for gravitational potential energy. Why can't it be ground? ground • (start) ~ ^ ^ = • m (later)