Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 6 – Work and Energy Chapter 6 • Work Done by a Constant Force • Kinetic Energy, and the Work-Energy Principle • Potential Energy • Conservative and Nonconservative Forces • Mechanical Energy and Its Conservation. Recalling Last Lecture Potential Energy Potential energy is the ability of an object to do work. Gravitational potential energy is (6-9) (6-10) Potential Energy The force needed to compress or stretch a string is written as: Where, k = spring stiffness constant Newton’s 3rd law says that the spring will exert a force FS in reaction to the force FP such that: (6-11) Eq. 6.11 (spring equation) is known as Hooke’s law FS is the force the spring exerts opposite to its displacement in an attempt to restore its original length Potential Energy Elastic potential energy is: (6-12) Conservative and NonNon-conservative Forces When the work done by a force does NOT depend on the path taken, this FORCE is said to be CONSERVATIVE. Example: 1) Gravitational force: F = mg; 2) Elastic force: F = -kx; W = mgh (h = vertical displacement) W = ½ kx 2 Note: If an object starts at a given point and returns to the same point, the net work done is ZERO if the forces acting on it are conservative. Example: You lift an object by a vertical displacement h: W1 = mgh You move the same back to its original position: W2 = mg(-h) = -mgh Net work done on the object: Wnet = W1 + W2 = mgh + (-mgh) = 0 Conservative and NonNon-conservative Forces When the work done by a force DOES depend on the path taken, this FORCE is said to be NON-CONSERVATIVE. Example: Friction forces the work done depends not only on the starting and ending points, but also on the path taken In general, forces that change as the direction of displacement changes are nonconservatives. Conservative and NonNon-conservative Forces We have seen that the potential energy associated to a force is the energy stored at some point. Gravitational potential energy is an example. This only makes sense if this energy can be uniquely defined at a given point (otherwise you would have two potential energies associate to a force at the same point two different abilities of an object at that particular point to do work this cannot be!). However, we just stated that the work done by NON-CONSERVATIVE forces do depend on the path taken It does NOT make sense to define potential energy associate to non-conservative forces as you would have more than one potential energy defined at a given point. Non-conservative forces do NOT have a potential energy associated to them. Thus, potential energy can only be defined for conservative forces. Conservative and NonNon-conservative Forces General form of the work-energy principle Observing the two previous slides, we can now generalize the work-energy principle to include the potential energy. Recalling: We previously wrote Wnet = ∆KE Now consider an object subject to both non-conservative and conservative forces. The net work done is: Where, Conservative and NonNon-conservative Forces General form of the work-energy principle But, from the work-energy principle, we know that: (6-13) We have also seen that the work done by a conservative force, WC , can be written in terms of the potential energy: We can then rewrite eq. 6-13 as: (6-14) Eq. 6-14 tells us that the work done by non-conservative forces is equal to the total change in kinetic and potential energies. Conservative and NonNon-conservative Forces Dissipative energy The work done by non-conservative force is NOT used to move an object The energy used to do this work is dissipated it will change the total mechanical energy in the system. Non-conservative forces are also known as DISSIPATIVE FORCES. Example: When I push my finger on a surface, the initial mechanical energy in my fingers is the sum of the potential and kinetic energies. But then friction acts against my finger’s motion. The work done by the friction force is not used to move your fingers. Instead, it tries to slow it down by dissipating part of the total energy available This energy is dissipated in the form of thermal energy (heat) you can feel your fingers getting warmer Mechanical Energy and its Conservation If ONLY conservatives forces are present , eq. 6.14 can be written as or (6-15) We can now define the total mechanical energy stored on a object at certain instant of time as: (6-16) And then rewrite eq. 6-15, yielding the equation of conservation of energy: (6-17) Mechanical Energy and its Conservation Eq. 6.17 states the following: “ If only conservative forces are acting, the TOTAL MECHANICAL ENERGY of a system neither increases nor decreases in any process. It stays constant it is CONSERVED” POWER: Power is the rate at which work is done, or the rate at which energy is transformed. (6-18) In the SI system the unit of power is: In general, if an object moves with average velocity 1 hp = 746 W , the power can be written as (6-19) Kinetic Energy, and the Work Energy Principle Problem 6-36 (textbook): In the high jump, Fran’s kinetic energy is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must Fran leave the ground in order to lift her center of mass 2.10 m and cross the bar with a speed of 0.70 m/s ? Kinetic Energy, and the Work Energy Principle Problem 6-36: We assume that all the forces on the jumper are conservative, so that the mechanical energy of the jumper is conserved. Subscript 1 represents the jumper at the bottom of the jump, and subscript 2 represents the jumper at the top of the jump. Call the ground the zero location for PE ( y = 0). We have: y1 = 0 y 2 = 2.10 m v 2 = 0.70 m s Solve for v1, the speed at the bottom. 1 2 m v12 + m gy1 = v1 = 1 2 m v 22 + m gy 2 v 22 + 2 gy 2 = → ( 0.70 m s ) 2 1 2 ( m v12 + 0 = + 2 9.80 m s 2 1 2 m v 22 + m gy 2 ) ( 2.10 m ) = → 6.45 m s Kinetic Energy, and the Work Energy Principle Problem 6-76 (textbook) An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot’s mass was 78 kg and his terminal velocity was 35 m/s, estimate (a) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; (c) the work done on him by air resistance as he fell. Kinetic Energy, and the Work Energy Principle Problem 6-76 : (a) Use conservation of energy, including the work done by the non-conservative force of the snow on the pilot. Subscript 1 represents the pilot at the top of the snowbank, and subscript 2 represents the pilot at the bottom of the crater. The bottom of the crater is the zero location for PE ( y = 0 ) . We have v1 = 35 m/s , y1 = 1.1 m, v2 = 0, y2 = 0 Solve for the non-conservative work: W N C + E1 = E 2 → W NC + W N C = − 12 m v12 − m gy1 = − 1 2 1 2 ( 78 m v12 + m gy1 = 1 2 m v 22 + m gy 2 ( → kg )( 35 m s ) − ( 78 kg ) 9.8 m s 2 = − 4.862 × 10 4 J ≈ − 4.9 × 10 4 J 2 ) (1.1 m ) Kinetic Energy, and the Work Energy Principle Problem 6-76 : (b) The work done by the snowbank is done by an upward force, while the pilot moves down. W N C = F sn o w d c o s 1 8 0 o = − F sn o w d Fsn o w = − W NC d = − − 4.862 × 10 4 J 1 .1 m → = 4.420 × 10 4 N ≈ 4.4 × 10 4 N Kinetic Energy, and the Work Energy Principle Problem 6-76 : (c) To find the work done by air friction, another non-conservative force, use energy conservation including the work done by the non-conservative force of air friction. Subscript 1 represents the pilot at the start of the descent, and subscript 2 represents the pilot at the top of the snowbank. The top of the snowbank is the zero location for PE ( y = 0 ) . We have v1 = 0 m/s , y1 = 370 m, v2 = 35 m/s, y2 = 0. Solve for the non-conservative work. W NC + E1 = E 2 W NC = 1 2 → W NC + 12 mv12 + mgy1 = mv 22 − mgy1 = 1 2 1 2 mv 22 + mgy 2 ( → ( 78 kg )( 35 m s ) − ( 78 kg ) 9.8 m s 2 = − 2.351 × 10 5 J ≈ − 2.4 × 10 5 J 2 ) ( 370 m ) Kinetic Energy, and the Work Energy Principle Problem 6-63 (textbook): A driver notices that her 1150-kg car slows down from 85 Km/h to 65 Km/h in about 6.0 s on the level when it is in neutral. Approximately what power (watts and hp) is needed to keep the car traveling at a constant 75 km/h? Kinetic Energy, and the Work Energy Principle Problem 6-63 The energy transfer from the engine must replace the lost kinetic energy. From the two speeds, calculate the average rate of loss in kinetic energy while in neutral. 1m s 1m s = 23.61 m s v = 65 km h = 18.06 m s 2 3.6 km h 3.6 km h 2 2 ∆ KE = 12 mv 22 − 12 mv12 = 12 (1150 kg ) (18.06 m s ) − ( 23.61 m s ) = − 1.330 × 10 5 J v1 = 85 km h P= W t = 1.330 × 10 5 J 6.0 s ( = 2.216 × 10 W , or 2.216 × 10 W 4 4 ) 746 W = 29.71 hp 1 hp Note now that 75 Km/h is the average between the car’s initial and final speeds: ⇒ We can use the 2.2 × 10 4 W or 3.0 × 101 hp is needed from the engine. to conclude