Chapter 7 – Kinetic energy, potential energy, work I. Kinetic energy. II. Work. III. Work - Kinetic energy theorem. IV. Work done by a constant force: Gravitational force V. Work done by a variable force. - Spring force. - General: 1D, 3D, Work-Kinetic Energy Theorem VI. Power VII. Potential energy Energy of configuration VIII. Work and potential energy IX. Conservative / Non-conservative forces X. Determining potential energy values: gravitational potential energy, elastic potential energy Energy: scalar quantity associated with a state (or condition) of one or more objects. I. Kinetic energy 1 Energy associated with the state of motion of an object. K mv2 (7.1) 2 Units: 1 Joule = 1J = 1 kgm2/s2 = N m II. Work Energy transferred “to” or “from” an object by means of a force acting on the object. To +W From -W - Constant force: Fx max v2 v2 v2 v2 2a d a 0 0 x x 2d 1 1 1 F ma m(v2 v2 ) ma d m(v2 v2 ) x x 2 0 d x 2 0 1 Work done by the force = Energy m(v2 v2 ) K K F d W F d 2 0 f i x x transfer due to the force. - To calculate the work done on an object by a force during a displacement, we use only the force component along the object’s displacement. The force component perpendicular to the displacement does zero work. F cos φ W Fxd F cos d F d (7.3) d - Assumptions: 1) F=cte, 2) Object particle-like. 90 W 180 90 W Units: 1 Joule = 1J = 1 kgm2/s2 90 0 A force does +W when it has a vector component in the same direction as the displacement, and –W when it has a vector component in the opposite direction. W=0 when it has no such vector component. Net work done by several forces = Sum of works done by individual forces. Calculation: 1) Wnet= W1+W2+W3+… 2) Fnet Wnet=Fnet d II. Work-Kinetic Energy Theorem K K f Ki W (7.4) Change in the kinetic energy of the particle = Net work done on the particle III. Work done by a constant force - Gravitational force: W F d mgd cos (7.5) Rising object: W= mgd cos180º = -mgd Fg transfers mgd energy from the object’s kinetic energy. Falling object: W= mgd cos 0º = +mgd Fg transfers mgd energy to the object’s kinetic energy. - External applied force + Gravitational force: K K f Ki Wa Wg (7.6) Object stationary before and after the lift: Wa+Wg=0 The applied force transfers the same amount of energy to the object as the gravitational force transfers from the object. IV. Work done by a variable force - Spring force: F kd (7.7) Hooke’s law k = spring constant measures spring’s stiffness. Units: N/m Hooke’s law 1D Fx kx Work done by a spring force: - Assumptions: • Spring is massless mspring << mblock • Ideal spring obeys Hooke’s law exactly. • Contact between the block and floor is frictionless. • Block is particle-like. Fx - Calculation: xi ∆x x f x 1) The block displacement must be divided into Fj many segments of infinitesimal width, ∆x. 2) F(x) ≈ cte within each short ∆x segment. x f x f Ws Fjx x 0 Ws F dx (kx) dx xi xi x f 1 2 x f 1 2 2 WS k x dx k x x k (x f xi ) x i i 2 2 1 2 1 2 W =0 If Block ends up at x =x . W k x k x s f i s 2 i 2 f 1 W k x2 if x 0 s 2 f i Work done by an applied force + spring force: K K f Ki Wa Ws Block stationary before and after the displacement: ∆K=0 Wa=-Ws The work done by the applied force displacing the block is the negative of the work done by the spring force. Work done by a general variable force: 1D-Analysis Wj Fj,avg x W Wj Fj,avg x better approximation more x, x 0 x f W F x W F(x)dx (7.10) lim j,avg x0 xi Geometrically: Work is the area between the curve F(x) and the x-axis. 3D-Analysis ˆ ˆ ˆ F Fxi Fy j Fzk ; Fx F(x), Fy F(y), Fz F(z) dr dx iˆ dy ˆj dz kˆ r x y z f f f f dW F dr Fxdx Fydy Fzdz W dW Fxdx Fydy Fzdz ri xi yi zi Work-Kinetic Energy Theorem - Variable force x f x f W F(x)dx ma dx xi xi dv dv ma dx m dx m v dx mvdv dt dx dv dv dx dv v dt dx dt dx v f v f 1 2 1 2 W mv dv m v dv mv f mvi K f Ki K 2 2 vi vi V. Power Time rate at which the applied force does work. - Average power: amount of work done in an amount of time ∆t by a force. W P (7.12) avg t - Instantaneous power: instantaneous time rate of doing work. dW P (7.13) F dt φ x dW F cos dx dx P F cos Fv cos F v (7.14) dt dt dt Units: 1Watt=1W=1J/s 1 kilowatt-hour = 1 kW·h = 3.60 x 106 J=3.6MJ 54. In the figure (a) below a 2N force is applied to a 4kg block at a downward angle θ as the block moves rightward through 1m across a frictionless floor. Find an expression for the speed vf at the end of that distance if the block’s initial velocity is: (a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in that the block is initially moving at 1m/s to the right, but now the 2N force is directed downward to the left. Find an expression for the speed of the block at the end of the 1m distance. W F d (F cos )d N F 2 2 x N W K 0.5m(v f v0 ) Fx 2 (a) v0 0 K 0.5mv f mg mg F Fy y 2 (2N)cos 0.5(4kg)v f v f cos m / s 2 2 2 (b) v0 1m / s K 0.5mv f 0.5(4kg)(1m / s) (c) v0 1m / s K 0.5mv f 2J 2 2 (2N)cos 0.5(4kg)v f 2J (2N)cos 0.5(4kg)v f 2J v 1 cos m / s f v f 1 cos m / s 18. In the figure below a horizontal force Fa of magnitude 20N is applied to a 3kg psychology book, as the book slides a distance of d=0.5m up a frictionless ramp. (a) During the displacement, what is the net work done on the book by Fa,the gravitational force on the book and the normal force on the book? (b) If the book has zero kinetic energy at the start of the displacement, what is the speed at the end of the displacement? N d W 0 y Only F , F do work x gx ax N (a) W WFa WFg or Wnet Fnet d Fgx x x F mg gy Fnet Fax Fg x 20cos30 mg sin 30 Wnet (17.32N 14.7N)0.5m 1.31J (b) K0 0 W K K f 2 W 1.31J 0.5mv f v f 0.93m / s 55. A 2kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at t=0, a steady wind pushes on the lunchbox in the negative direction of x, Fig. below. Estimate the kinetic energy of the lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force from the wind do on the lunch box from t=1s to t=5s? Motion concave downward parabola 1 x t(10 t) 10 dx 2 v 1 t dt 10 dv 2 a 0.2m / s2 dt 10 F cte ma (2kg)(0.2m / s2 ) 0.4N (b) t 5s v f 0 W F x (0.4N)(t 0.1t 2 ) K f 0J (a) t 1s v f 0.8m / s (c) W K K f (5s) K f (1s) K 0.5(2kg)(0.8m / s)2 0.64J f W 0 0.64 0.64J 74. (a) Find the work done on the particle by the force represented in the graph below as the particle moves from x=1 to x=3m. (b) The curve is given by F=a/x2, with a=9Nm2. Calculate the work using integration (a) W Area under curve W (11.5squares)(0.5m)(1N) 5.75J 3 3 9 1 1 (b) W dx 9 9( 1) 6J 2 1 x x1 3 73. An elevator has a mass of 4500kg and can carry a maximum load of 1800kg. If the cab is moving upward at full load at 3.8m/s, what power is required of the force moving the cab to maintain that speed? F m 4500kg 1800kg 6300kg a total P F v (61.74kN)(3.8m / s) Fa mg Fnet 0 Fa Fg 0 P 234.61kW 2 Fa mg (6300kg)(9.8m / s ) 61.74kN mg A single force acts on a body that moves along an x-axis.