Physics 2210 Fall 2015 smartphysics 07 Kinetic Energy and Work (continued) 09/25/2015
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Unit 07 Kinetic Energy and work: 1d Definition: kinetic Energy: K 1 2 mv2 units: joule (J) = kg m 2 /s 2 Definition: work W j done by (vector) force F i on object: W j x i x f F j Definition: NET (total) work W done (by net force F) on object: W W 1 + W 1 + + W N x i x f F x f x f x f F 1 + F 2 + + F N x i x i x i Work-(Kinetic) Energy Theorem (proof to come) K = W In moving from position x i to x f while under the influence of forces F 1, F 2, F N, the change in the kinetic energy of an object is equal to the net (total) work done by all the forces acting on it.
Special case: if the force F j is constant, then the work it does over the object as the x f latter moves from x 1 to x 2 is then W j F j = F x i j = F x i j x = F j x 2 x 1 x f
Poll 09-23-02 A box sits on the horizontal bed of a truck accelerating to the left as shown. Static friction between the box and the truck keeps the box from sliding around as the truck drives. The work done on the box by the static frictional force as the accelerating truck moves a distance D to the left is? A. Depends on the speed of the truck. B. Zero C. Positive D. Negative
Work-Energy Theorem in 1d Definition: kinetic Energy of an object (from Pre-lecture) K 1 2 mv2 Newton co-developed calculus, so let s take the time-derivative = 1 d v2 2m = 1 dv 2m 2v = mmm = v mm = vv where v is the velocity of the object, and F the net force on m Integrating over time from t i to t f K 1 2 mv f 2 1 t f t f 2 mv i 2 = vv = F v t i t i t f = F x f = F t i x i Remember F is the net force on m F = F 1 + F 2 + + F N, so x f K 1 2 mv f 2 1 2 mv i 2 = F 1 + F 2 + + F N x i x i x f x i x f
Example 7-1 (1/4) A 4.1-kg particle moving along the x axis has a velocity of +1.9 m/s as it passes through the origin. It is subjected to a single force, F x, that varies with position, as shown. Find (a) the kinetic energy of the particle as it passes through the origin? (b) the work is done by the force as the particle moves from x = 0.0 m to x = 4.0 m (c) the speed of the particle when it is at x = 4.0 m? (%i1) /* kinetic energy is given by KE=0.5*m*v^2 */ KE: 0.5*m*v^2; 2 (%o1) 0.5 m v (%i2) /* at origin */ KE0: KE, v=v0; 2 (%o2) 0.5 m v0 (%i3) /* put in numbers: m=4.1kg, v=v0=+1.9m/s */ KE0, m=4.1, v0=1.9; (%o3) 7.400499999999999 Answer (a) 7.40 J
Example 7-1 (2/4) A 4.1-kg particle, with v = +1.9 m/s passing the origin. It is subjected to F x, that varies with position, as shown. Find (a) the kinetic energy of the particle as it passes through the origin? (b) the work is done by the force as the particle moves from x = 0.0 m to x = 4.0 m (c) the speed of the particle when it is at x = 4.0 m? /* Work done is the area under the Fx vs x curve so from x=0 to x=4m we have a triangle of base 4m and hieght 6N --> Area=W=0.5*(4m)*(6N) = 12J */ W: base * height; (%o4) base height (%i5) W, base=4, height=6; (%o5) 24 Answer (b) 24J (%i7) /* alternatively, we can write Fx piece wise For x=0 to x=2m we have Fx1=3*x */ Fx1: 3*x; (%o7) 3 x (%i9) W1x: integrate(fx1, x); 2 3 x (%o9) ---- 2... continued
Example 7-1 (3/4) A 4.1-kg particle, with v = +1.9 m/s passing the origin. It is subjected to F x, that varies with position, as shown. Find (a) the kinetic energy of the particle as it passes through the origin? (b) the work is done by the force as the particle moves from x = 0.0 m to x = 4.0 m (c) the speed of the particle when it is at x = 4.0 m? (%i11) W1: integrate(fx1, x, 0, 2); (%o11) 6 (%i15) /* from x=2 to x=4, we have Fx=-3*(x-4)... note that the x=intercept is at x=4 and slope is -3 */ Fx2: -3*(x-4); (%o15) - 3 (x - 4) (%i16) W2x: integrate(fx2, x); 2 x (%o16) - 3 (-- - 4 x) 2 (%i17) W2: integrate(fx2, x, 2, 4); (%o17) 6 (%i18)w1 + W2; (%o18) 12 (b) Same Answer: 12 J
Example 7-1 (4/4) A 4.1-kg particle, with v = +1.9 m/s passing the origin. It is subjected to F x, that varies with position, as shown. Find (a) (b) the kinetic energy of the particle as it passes through the origin? the work is done by the force as the particle moves from x = 0.0 m to x = 4.0 m (c) the speed of the particle when it is at x = 4.0 m? (%i20) /* Final kinetic energy is KE4=KE0+W */ KE4: KE0 + W, m=4.1, v0=1.9, W=W1+W2; (%o20) 19.4005 **** restarted maxima here... (%i1) eqn1: KE=0.5*m*v^2; 2 (%o1) KE = 0.5 m v (%i2) soln1: solve(eqn1, v); rat: replaced -0.5 by -1/2 = -0.5 KE KE (%o2) [v = - sqrt(2) sqrt(--), v = sqrt(2) sqrt(--)] m m (%i3) /* take positive root */ v4: rhs(soln1[2]), numer; KE 0.5 (%o3) 1.414213562373095 (--) m (%i4) v4, KE=19.4005, m=4.1; (%o4) 3.076305988777021 Answer(c) 3.08 m/s
Poll 09-25-01 A car drives up a hill with constant speed. Which statement best describes the total (net) work done on the car by all forces as it moves up the hill with constant speed? A. Positive B. Zero C. Negative
Extending it all to more than 1d Kinetic Energy: v 2 v v, K 1 2 mv2 = 1 m v 2 = 1 mv v 2 2 Work: F j F j dr, W j F j dr (Math Detour) Vector Algebra Definition: Because of the animation in this lecture, the Powerpoint (PPTX) file for this lecture is also posted on CANVAS along with the PDF dot product, scalar product or inner product A B is the product of A B, the component of A along the direction of B, with the magnitude B of B: A B A B B = A ccc θ A,B B = A B ccc θ A,B Where θ A,B is the angle between the directions of A and of B. A θ A,B A B = A cos θ A,B B Note that the cosine of an acute angle is positive And cosine of an obtuse angle is negative: A B = A cos θ A,B > 0 for θ A,B < 90 < 0 for θ A,B > 90 For aitional help on dot product, see Khan Academy, for example: https://www.khanacademy.org/science/physics/electricity-magnetism/electric-motors/v/the-dot-product
Kinetic Energy in more than 1d Kinetic Energy: v 2 v v, K 1 2 mv2 = 1 m v 2 = 1 mv v 2 2 Work: F j F j dr, W j F j dr Dot Product continued Note that the dot product is commutative ( abelian ) i.e. B A = A B B A B A A = B cos θ A,B A NOTE: Since cos 90 = A B cos θ A,B = A B = 0, we have A B = B A = 0 if A B Because of the animation in this lecture, the Powerpoint (PPTX) file for this lecture is also posted on CANVAS along with the PDF A θ A,B B A = B cos θ A,B B Note that the cosine of an acute angle is positive And cosine of an obtuse angle is negative: A B = B A > 0 for θ A,B < 90 < 0 for θ A,B > 90
Dot Product and Cartesian Components A y = A ȷ = A cos φ A 90 = A sin φ A A = A x ı + A y ȷ Note we used the trig. Identity: cos φ 90 A φ A A x = A ı = A cos φ A cos 90 φ sin φ Also: ı ı = ȷ ȷ = 1 1 cos 0 = 1 and ı ȷ = ȷ ı = 1 1 cos 90 = 0 AND Dot Product is distributive and so: A B = A x ı + A y ȷ B x ı + B y ȷ = A x B x ı ı + A x B y ı j + A y B x ȷ ı + A y B y ȷ ȷ = A x B x + A y B y (+A z B z if 3d) which is a VERY NICE AND CONVENIENT formula for working on computers For example v 2 = v v = v x v x + v y v y (+v z v z if 3d) ȷ y ı B = B x ı + B y ȷ B φ B B x = B ı = B cos φ B B y = B ȷ = B cos φ B 90 = B sin φ B x
Change in Kinetic Energy Definition of kinetic Energy from Pre-lecture K 1 2 mv2 = 1 2 mv v = 1 2 m v x 2 + v y 2 Newton co-developed calculus, so let s take the time-derivative = 1 2 m d v x 2 + d v y 2 Let s take a look at the first term inside the bracket d v x 2 = v x dv x + dv x v x = 2v x dv x = 2 a x Similarly for the second term: d v 2 y = 2 dy a y = 1 2 m d v x 2 + d v y 2 = ma x + ma y dy Integrating over time from t i to t f K 1 2 mv f 2 1 2 mv i 2 = t i t f ma x + ma y
Change in Kinetic Energy Definition of kinetic Energy from Pre-lecture K = 1 2 m v x 2 + v y 2 From previous slide K 1 2 mv f 2 1 2 mv i 2 = ma x + ma y t i Newton s 2 nd Law (N2L): a x = F x /m, and a y = F y /m, Where F x, F y are the x- and y-components of the NET force on the object t f r K = F x + F f y = F t i x + F y K 1 2 mv f 2 1 2 mv i 2 = F nnn dr This last integral is called a path integral or line integral. This is the subject matter of most Calc-III classes. You need to understand what a path integral means but you won t be asked to actually DO one in PHYS2210 (you will see it in later physics and engineering courses). t f
Path from to, e.g. path taken by a barge along a river Vectors representing a variable force that changes along the path e.g. a force exerted by A horse pulling a barge from shore of a river Definition of Path/Line Integral Approximate path by short, straight segments r n r n... F r n r N r N = F x x n, y n ı + F y x n, y n ȷ r n = x n ı + y n ȷ r 3 r 1 r 2 r 1 r r 3 2 F dr... Take limit of infinitesimal segments N lim F r N n n=1 Sum the dot products over the segments r n r n = x n, y n = x n ı + y n ȷ Take dot product F r n F r n r n= Fx x n, y n x n + F y x n, y n y n
Path from to, e.g. path taken by a barge along a river Vectors representing a variable force that changes along the path e.g. a force exerted by A horse pulling a barge from shore of a river W : WORK done by a force F Approximate path by short, straight segments r n r n... F r n r N r N = F x x n, y n ı + F y x n, y n ȷ r n = x n ı + y n ȷ r 3 r 1 r 2 r 1 r r 3 2... Take limit of infinitesimal segments Sum the dot products over the segments r n = x n, y n = x n ı + y n ȷ Take dot product F r n W = F dr N lim n F r n n=1 r n F r n r n= Fx x n, y n x n + F y x n, y n y n
Only components (of the forces) parallel to the path/trajectory of motion does any work. Again, parallel force components in the direction of motion does POSITIVE work (e.g. mm cos θ) Parallel force components opposite direction of motion does NEGATIVE work (e.g. friction force f) Components of forces perpendicular to the path of motion does ZERO work Type equation here. Path of motion Work fricton
Work done by a constant force In general the work line integral is a nightmare In some special cases the integral simplifies Case 1: Constant Force (it is in this case sufficient that the force is constant along the path taken by the object) W = F dr = F x + F y = F x + F y r i This factorization AND separation of the dx- and dy-integrals is in general not possible.!!! It is allowed here because F x and F y are constant along the path of integration W = F x x + F y y = F r
WORK done by F that is constant along the path Path from to, e.g. path taken by a barge along a river F = F x ı + F y ȷ Constant force r n = x n ı + y n ȷ Take limit of infinitesimal segments Sum the dot products over the segments r n = x n, y n = x n ı + y n ȷ Take dot product F r n F r n = F x x n + F y y n F dr N lim F r n n n=1 = lim n N F r n n=1
WORK done by F that is constant along the path Path from to, e.g. path taken by a barge along a river Constant force F = F x ı + F y ȷ S r = xı + yȷ S = r F = F x ı + F y ȷ θ F = F cos θ W = F r = F x x + F y y W = F r = F S = FS cos θ
Poll 09-25-02 Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. In which case does the object have the biggest total (net) work done on it by all forces during its motion? A. Same B. Free Fall C. Incline D. String
My Notation F dr F dl smartphysics Notation Equivalent