Physics : Lecture 10, Pg 1 Engineering Physics : Lecture 10 (Chapter 7 Halliday) l Work done by variable force è Spring l Problem involving spring & friction l Work done by variable force in 3-D è Newton’s gravitational force

Physics : Lecture 10, Pg 2 Review: Constant Force... F  W = F  d No work done if  = 90 o. T è No work done by T. N è No work done by N. v N T v

Physics : Lecture 10, Pg 3 Review: Work/Kinetic Energy Theorem: NetWork {Net Work done on object} = changekinetic energy {change in kinetic energy of object} W F =  K = 1 / 2 mv / 2 mv 1 2 xxxx F v1v1 v2v2 m W F = F  x

Physics : Lecture 10, Pg 4 Work done by Variable Force: (1D) When the force was constant, we wrote W = F  x è area under F vs. x plot: l For variable force, we find the area by integrating: è dW = F(x) dx. F x WgWg xx F(x) x1x1 x2x2 dx

Physics : Lecture 10, Pg 5 Work/Kinetic Energy Theorem for a Variable Force F F dx dv dx dv v dv v22v22 v12v12 v22v22 v12v12 dv dx v dv dx v(chain rule) dt = =

Physics : Lecture 10, Pg 6 1-D Variable Force Example: Spring l For a spring we know that F x = -kx. F(x) x2x2 x x1x1 -kx relaxed position F = - k x 1 F = - k x 2

Physics : Lecture 10, Pg 7 Spring... l The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2. WsWs F(x) x2x2 x x1x1 -kx relaxed position

Physics : Lecture 10, Pg 8 Spring... F(x) x2x2 WsWs x x1x1 -kx l The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2.

Physics : Lecture 10, Pg 9 Work & Energy l 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. è If the initial speed of the box were doubled and its mass were halved, how far x 2 would the spring compress ? x (a) (b) (c) (a)  (b) (c)

Physics : Lecture 10, Pg 10 Solution Again, use the fact that W NET =  K. x1x1 v1v1 so kx 2 = mv 2 m1m1 m1m1 In this case, W NET = W SPRING = - 1 / 2 kx 2 and  K = - 1 / 2 mv 2 In the case of x 1

Physics : Lecture 10, Pg 11 Solution x2x2 v2v2 m2m2 m2m2 So if v 2 = 2v 1 and m 2 = m 1 /2

Physics : Lecture 10, Pg 12 Problem: Spring pulls on mass. l A spring (constant k) is stretched a distance d, and a mass m is hooked to its end. The mass is released (from rest). What is the speed of the mass when it returns to the relaxed position if it slides without friction? relaxed position stretched position (at rest) d after release back at relaxed position vrvr v m m m m

Physics : Lecture 10, Pg 13 Problem: Spring pulls on mass. l First find the net work done on the mass during the motion from x = d to x = 0 (only due to the spring): stretched position (at rest) d relaxed position vrvr m m i

Physics : Lecture 10, Pg 14 Problem: Spring pulls on mass. l Now find the change in kinetic energy of the mass: stretched position (at rest) d relaxed position vrvr m m i

Physics : Lecture 10, Pg 15 Problem: Spring pulls on mass. Now use work kinetic-energy theorem: W net = W S =  K. stretched position (at rest) d relaxed position vrvr m m i

Physics : Lecture 10, Pg 16 Problem: Spring pulls on mass. Now suppose there is a coefficient of friction  between the block and the floor f Δr The total work done on the block is now the sum of the work done by the spring W S (same as before) and the work done by friction W f. W f = f. Δr = -  mg d stretched position (at rest) d relaxed position vrvr m m i f =  mg r r r r

Physics : Lecture 10, Pg 17 Problem: Spring pulls on mass. Again use W net = W S + W f =  K W f = -  mg d stretched position (at rest) d relaxed position vrvr m m i f =  mg r r r r

Physics : Lecture 10, Pg 18 Work by variable force in 3-D: F l Work dW F of a force F acting through an infinitesimal dr displacement dr is: Fdr dW = F. dr l The work of a big displacement through a variable force will be the integral of a set of infinitesimal displacements: Fdr W TOT = F. drF drdrdrdr 

Physics : Lecture 10, Pg 19 Work by variable force in 3-D: Newton’s Gravitational Force dr l Work dW g done on an object by gravity in a displacement dr is given by: Fdr rr rrr dW g = F g. dr = (-GMm / R 2 r). (dR r + Rd  ) dW g = (-GMm / R 2 ) dR (since r.  = 0, r. r = 1) ^ ^ ^ r  ^ ^ dr Rd  dR R FFgFFg m M dd ^ ^ ^^

Physics : Lecture 10, Pg 20 Work by variable force in 3-D: Newton’s Gravitational Force l Integrate dW g to find the total work done by gravity in a “big” displacement: W g = dW g = (-GMm / R 2 ) dR = GMm (1/R 2 - 1/R 1 ) F F g (R 1 ) R1R1 R2R2 F F g (R 2 ) R1R1 R2R2 R1R1 R2R2 m M

Physics : Lecture 10, Pg 21 Work by variable force in 3-D: Newton’s Gravitational Force not on the path taken l Work done depends only on R 1 and R 2, not on the path taken. R1R1 R2R2 m M

Physics : Lecture 10, Pg 22 Newton’s Gravitational Force Near the Earth’s Surface: Suppose R 1 = R E and R 2 = R E +  y but we have learned that So: W g = -mg  y R E +  y M m RERE

Physics : Lecture 10, Pg 23 Recap of today’s lecture l Review l Work done by gravity near the Earth’s surface l Examples: è pendulum, inclined plane, free fall l Work done by variable force è Spring l Problem involving spring & friction l Work done by variable force in 3-D è Newton’s gravitational force Look at textbook problems l Look at textbook problems