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Transcript
Chapter 6 Work and kinetic energy In chapters 4 and 5 we studied Newton’s laws of motion and applied them to various situations. In chapter 6 we shall introduce a new approach that makes the solution of mechanics problem easier The following concepts will be introduced • Kinetic energy (symbol: K ) • Work (symbol: W ) We will also introduce and use the work-energy theorem (6-1) The scalar product of two vectors (also known as "dot" product) A B AB cos Note 1: A B B A Note 2: A B 0 when 90 The dot product in terms of vector components A Ax i Ay j Az k B Bx i By j Bz k A B Ax Bx Ay By Az Bz (6-2) Work-energy theorem (6-3) Consider the motion of an object of mass m along the x-axis from point 1 (coordinate x1) to point 2 (coordinate x2). A constant net force Fnet acts on the object. The velocity at points 1 and 2 is v1 and v2 , respectively a v1 . O . 1 x1 2a( x2 x1 ) v22 v12 m Fnet v2 . 2 x2 x-axis (third equation of kinematics) Fnet a Substitute a in the equation above m 2 2 Fnet mv mv 2 2 ( x2 x1 ) v22 v12 Fnet ( x2 x1 ) 1 m 2 2 v1 . O . m 1 x1 x Fnet v2 (6-4) . 2 x2 x-axis mv22 mv12 Fnet ( x2 x1 ) Work-energy theorem x2 x1 x 2 2 mv22 mv12 Fnet x We define as work (symbol W) 2 2 performed by Fnet during the motion from point 1 to point 2: Wnet Fnet x Units: Nm Joule (J) We define as the kinetic energy of a moving object kg m 2 Units: Nm J 2 s The work-energy theorem can be written as : mv 2 K 2 Wnet K Algebraic sign for Wnet = Fnetx (all possible scenarios) (6-5) I II III IV . O . O . O . O . 1 . 2 . 1 . 2 m Fnet x m x Fnet . 2 Fnet . x-axis 1 x-axis m Fnet m x x . x > 0 Fnet > 0 Wnet > 0 x < 0 Fnet > 0 Wnet < 0 x-axis x > 0 Fnet < 0 Wnet < 0 x-axis x < 0 Fnet < 0 Wnet > 0 2 . 1 Caution If there are more than one forces (F1 , F2, F3 , …FN) acting on the moving object we calculate the net work Wnet as follows: We first determine the work each force performs: W1 = F1 x , W2 = F2 x , W3 = F3 x , …, WN = FN x Wnet is simply the sum of all the terms above i.e. Wnet = W1 + W2 + W3 + …+ WN Note: The work-energy theorem Wnet = K applies for the net work. All forces acting on the object for which we are applying the theorem must be included (6-6) Example (6-1) page 146 A car of mass m = 1200 kg falls a vertical distance h = 24 m starting from rest. Find the velocity v2 of the car before it hits the water. .1 Fnet mg Wnet Fnet y (-mg )(-h) mgh y Wnet 1200 9.8 24 2.8 105 J mg Wnet K y h .2 water (6-7) v2 (work-energy theorem) mv22 mv12 K v1 0 2 2 2 mv2 Wnet v2 2 gh 2 v2 2 9.8 24 21.7 m/s Work-energy theorem when we have (6-8) motion of an object of mass m in a plane m from point 1 to point 2 under the action of a constant force F Displacement = r Work-energy theorem: Wnet K Below we give the general definition of W W F r F r cos mv22 mv12 K v1 and v2 are the velocities of the object 2 2 at point 1 and point 2, respectively 1 m mg r An object of mass m is thrown with an initial speed v1 off a tall building. Find the object’s speed after it has fallen a vertical distance h Wnet h 2 L mv22 mv12 K 2 2 r Li h j Fnet mg j A B Ax Bx Ay By Wnet Fnet r ( mg )( h) mgh mv22 mv12 K mgh 2 2 (6-9) v2 v12 2 gh Solve for v 2 (6-10) Consider the motion of an object from point xo to point xf along the xaxis under the action of a force F(x) that is not constant. xf In this case the work W performed by F is given by: W F ( x)dx xo W Area under the F versus x curve from x o to x f Work performed by a spring (spring constant = k) as it is stretched from its relaxed length by L L x L2 W F ( x)dx (kx)dx k xdx k k 2 2 0 0 0 0 L L L 2 W = shaded area in the F versus x plot (6-11) The most general case is when the force F changes both magnitude and direction from point to point. In this case the work W performed by F as it moves an object from point A to point B along a given path. W depends on rA, rB and the path. W is calculated as follows: (6-12) • Divide the path into segments r • Calculate the work W = F•r for each element • Sum all the contributions and take the limit as r 0 rB The limit of the sum gives: W F d r rA This type of integral is known as "line intergal" (6-13) Work performed during uniform circular motion. The net force F points towards the center C of the orbit (centripetal force). For the path segment ds the work dW is: dW F d s Fds cos(90) 0 W dW 0 Conclusion: C . No work is done on an object that undergoes uniform circular motion (6-14) y A Classification of forces 1 2 B 3 Work W performed by a force F as it moves an object along one of the three paths from point A to point B: rB O x W F dr rA A force is called “conservative” if W does not depend on the path but only on the coordinates of the start and finish points. In this case: W1 = W2 = W3 A force is called “non-conservative” if W depends not only on the coordinates of the start and finish points but on the path as well. In this case W1 W2 W3 (6-15) y A (6-16) 1 C B 2 D x O If the force F is conservative than W along any closed path is zero. This statement can be used as an alternative definition of a conservative force Consider a closed path ACBDA. This can be divided into two different paths that take us from point A to point B. Path 1 (ACB), and path 2 (ADB). WACB = WADB A WACBDA WACB WBDA WBDA F d r (along path 2 ) B A B WBDA F d r F d r WADB B A WACBDA WACB WADB 0 Example of a conservative force: The gravitational force We shall prove that the work gone by Fg along path 1 and path 2 is the same. Path 1: W1 = Fg•r1 W1 = mgLcos(90-) cos (90 - ) = sin Path 1 r1 W1 = mgLsin Path 2: W2 = WAC + WCB WAC = mghcos0 = mgh h = Lsin WAC = mgLsin Path 2 WCB = mgLcos(90) = 0 W2 = mgLsin = W1 (6-17) (6-18) . A . B Example of a non-conservative force: friction f We shall calculate the work the work done by friction as it moves the cup along a closed path that starts at point A and ends at point A. During the trip we apply a force F = -f so that the net force on the cup, and thus its acceleration a is zero r1 A f1 . r2 f2 . B (6-19) f1 = f2 = kmg Note: friction opposes motion W = WAB + WBA WAB = f1 xmaxcos(180) = - f1xmax = -kmgxmax WBA = f2 xmaxcos(180) = - f2xmax = -kmgxmax W = -kmgxmax -kmgxmax = -2kmgxmax 0 Power Consider a force F that moves an object of mass m from point x to point x + dx in time dt F . O . t m x (6-20) t+dt . x+dx x-axis dx Power (symbol P) is defined as the rate at which F performs work. dW P dt Units: Joule/second Watt (W) dW Fdx dx dW Fdx P F Fv dt dt dt Note: The equation P = Fv is valid when F and the displacement dx are parallel y (6-21) P dr v path O F Fx i Fy j In general the force F and the displacement dr are not parallel. In this case the power P is given by the equation: F P F v Fv cos x , d r dxi dy j dW F d r Fx dx Fy dy dW Fx dx Fy dy P Fx vx Fy v y F v dt dt Commonly used practical (non-SI) Units P = W/t W = Pt Power: The horsepower (symbol: hp) is the average power that a horse can generate 1 hp = 746 W Work: The kilowatt-hour (symbol: kWh) is the work produced by a machine of power P = 1 kW = 1000 W in a time interval t = 1 hour = 3600 s W = Pt 1 kWh = 10003000 Ws = 3.6106 J (6-22)