Physics 2210 Fall smartphysics 09 Work and Potential Energy, Part II 10/02/2015

Transcription

1 Physics 10 Fall 015 smartphysics 09 Work and Potential Energy, Part II 10/0/015

2 r i Top View O In the limit N the blue path becomes the black path Frictionless, horizontal surface Spring of force constant k relaxed length L 0 W s = The green path gives the same work as the blue path r f m r f The red path also gives the same work as the green and blue paths Path from r i to r f Approximate path by N that alternate n = 1,,3, N, (n 1)th: radial (n)th: arc F s r for arcs: work done along the even segments vanish F r n r n = 0 F s r for radial steps: Only work done along the odd segments contribute F r n 1 r n 1 = F r n 1 r n 1 F(r)dd lim N F r n 1 r n 1 r i n=1 N F r = k(r L 0 )

3 r f W s F s r dd r i Work done by a Spring r f = k r L 0 dd Change of Variable x = r L 0, x = dr r i r f L 0 W s k xdx = 1 kx r i L 0 r i L 0 W s = 1 k r f L 0 ri L 0 r f = k r L 0 dd r i r f L 0 Or of we let x represent the deformation (which we already did), then W s = 1 k L f Li, or W s = 1 k x f x i Potential Energy of a Spring U S r U S L 0 = W s L0 r = 1 k r L 0 L 0 L 0 = + 1 k r L 0 0 = 1 k r L 0 Taking the relaxed spring to have ZERO potential energy (usually best choice) NOTE: U S = 1 k L, or U S = 1 kx x is not spring length NOTE: x is not spring length

4 Poll A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 1 from its relaxed position while momentarily coming to rest. If the initial speed of the box were doubled, how far x would the spring compress? A. x = x 1 B. x = x 1 C. x = 4 x 1 NOTE: x is not spring length

5 Example 08-0 (1/3) A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring is then used to project a second block of mass 4m, giving it a speed of 5v. What distance x was the spring compressed in the second case? Answer in terms of a numerical factor times x, the compression of the first block. (%i1) /* total energy */ Energy: 0.5*mass*speed^ + 0.5*k*compression^; (%o1) 0.5 mass speed compression k (%i) E1i: Energy, mass=m, speed=0, compression=x; (%o) 0.5 k x (%i3) E1f: Energy, mass=m, speed=v, compression=0; (%o3) 0.5 m v (%i4) eqn1: E1i = E1f; (%o4) 0.5 k x = 0.5 m v (%i5) Ei: Energy, mass=4*m, speed=0, compression=x; (%o5) 0.5 k x... continued NOTE: x is not spring length

6 Example 08-0 (/3) Spring on mass m, compression x, results in speed v. same spring on mass 4m: speed of 5v. What distance was the spring compressed in the second case? i.e. x / x =? (%i6) Ef: Energy, mass=4*m, speed=5*v, compression=0; (%o6) 50.0 m v (%i7) eqn: Ei = Ef; (%o7) 0.5 k x = 50.0 m v (%i8) /* strategy: solve for k in each case */ soln1: solve(eqn1, k); m v (%o8) [k = ----] x (%i9) k1: rhs(soln1[1]), numer; m v (%o9) ---- x NOTE: x is not spring length... continued

7 Example 08-0 (3/3) Spring on mass m, compression x, results in speed v. same spring on mass 4m: speed of 5v. What distance was the spring compressed in the second case? i.e. x / x =? (%i10) soln: solve(eqn, k); 100 m v (%o10) [k = ] x (%i11) k: rhs(soln[1]), numer; 100 m v (%o11) x (%i1) soln3: solve(k1=k, x); (%o1) [x = - 10 x, x = 10 x] (%i13) /* take positive root */ x: rhs(soln3[]); (%o13) 10 x Answer: 10 NOTE: x is not spring length

8 Universal Gravitation (in HW for unit 9) r M F G m We treat the gravitational force exerted by a very large spherical mass M on a small mass m as if M is stationary with its center at the origin. Then the force on m always points towards the origin, and with a radial component of F r = GGG r The minus sign means it points inward r f W G = F r r dd r i r f = r i GGG r dd W G = GGG r f = GGG r dd r i 1 r f 1 r i GGm = 1 r 1 r i r f It is conventional to choose U G = 0 at r = U G = 0 GGG 1 r 1 = GGG r U G = GGG r

9 Poll (checkpoint for unit 8) Consider two identical objects released from rest high above the surface of the earth (neglect air resistance for this question). Case 1: we release an object from a height above the surface of the earth equal to 1 earth radius, its kinetic energy just before it hits the earth is K 1. Case we release an object from a height above the surface of the earth equal to earth radii, its kinetic energy just before it hits the earth to be K. Compare the kinetic energy of the two just before they hit the surface of the earth. A. K = K 1 B. K = 4 K 1 C. K = (4/3) K 1 D. K = (3/) K 1

10 Unit 09 In your instructor terms: E = j work done by non conservative force j

11 Poll A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance D from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is μ k. What is the macroscopic work done on the block by friction during this process? A. mgh B. mgh C. μ k mgd D. 0

12 Example (1/3) The kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i1) Energy: 0.5*m*v^ + m*g*y; (%o1) g m y m v (%i) /* at top from rest */ E0: Energy, v=0, y=h; (%o) g h m (%i3) /* at bottom of ramp v=v1, y=0 */ E1: Energy, v=v1, y=0; (%o3) 0.5 m v1 (%i4) /* Energy is conserved down the ramp: no friction Solve for v1 */ soln1: solve(e1=e0, v1); rat: replaced 0.5 by 1/ = 0.5 (%o4) [v1 = - sqrt() sqrt(g h), v1 = sqrt() sqrt(g h)]

13 Example (/3) The kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i5) /* take positive root */ v1: rhs(soln1[]); (%o5) sqrt() sqrt(g h) (%i7) v1, g=9.81, h=4, numer; (%o7) Answer (a) 8.86 m/s (%i8) /* Energy dissipated by friction is the total energy of the system before the rough surface */ E0, m=, g=9.81, h=4; (%o8) Answer (b) 78.5 J (%i9) /* Work done by surface is W=-E0, but it is also W=-Fs*d where Fs is the magnitude of the friction force which is given by Fs=mu_k*N, but here the normal force has magnitude N=m*g */ N: m*g; (%o9) g m... continued

14 Example (3/3) The kg block slides down a frictionless curved ramp, starting from rest at a height of h = 4 m. The block then slides d = 10 m on a rough horizontal surface before coming to rest. (a)what is the speed of the block at the bottom of the ramp? (b)what is the energy dissipated by friction? (c) What is the coefficient of friction between the block and the horizontal surface? (%i10) Fs: mu_k*n; (%o10) (%i11) soln: solve(-fs*d=-e0, mu_k); g m mu_k (%o11) [mu_k = -] (%i1) mu_k: rhs(soln[1]); (%o1) - (%i13) mu_k, h=4, d=10; (%o13) - Answer (c) mu_k = 0.40 h d 5 h d

15 Unit 09

16 Poll The potential energy of an object U as a function of x looks like the plot shown above. Where is the force the biggest in the negative x direction? a b c d

17 Unit 10

18 Unit 10

19 Poll A glazed doughnut is shown above. Where is the center of mass of this doughnut? A. The center of mass is not defined in cases where mass is missing. B. Somewhere inside the solid part of the doughnut C. In the center of the hole.