How to calculate work done by a varying force along a curved path. The meaning and calculation of power in a physical situation

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1 Chapter 6: Work and Kinetic Energy What is work done by a force What is kinetic energy work-energy theorem How to calculate work done by a varying force along a curved path The meaning and calculation of power in a physical situation

2 What does the work do (mean): The work-energy theorem Work done on an object can change its motion and energy. Relate these three topics we ve studied so far: Newton s second law: F=ma Definition of work: W=Fs Straight line motion (constant acceleration) v 22 =v 12 +2as W = Fs = 1 2 mv mv 2 1 Definition of kinetic energy: K = 1 2 mv 2 In later chapter, this is related to energy conservation! Work-energy theorem: W tot = K 2 K 1 = ΔK work done by the net force equals the change in kinetic energy

3 Work, a force through a distance As in the illustration, the pushing (force) is in the same direction that the object moves The work is only done by a force when the force contributes to the motion What if F and s are not in the same direction? Unit: joule 1J =1N m

4 Use the parallel component if the force acts at an angle Only the component parallel to the displacement does the work W = F s = Fscosφ φ: the angle between If the force and displacement are known by their compoenents F = F xˆ i + F ˆ y j = (160N) i ˆ (40N) ˆ j s = xˆ i + yˆ j = (14m)ˆ i + (11m) ˆ j then W = F x x + F y y F, s Scalar product Example 6.1 (b)

5 Scalar product of two vectors Numbers can be multiplied; What about vectors It can be multiplied: Scalar product and cross product Dot product (Available in CH1 vectors in the text book) A B = ABcosφ = A B cosφ What else do you need to know? What if φ=90 o? φ=0 o? φ=180 o Using components (2d) A B = A x B x + A y B y It is a Scalar, and has no direction! A B = B A i ˆ i ˆ =1 i ˆ ˆ j = 0

6 How to find angles between two vectors: from 2d to 3d How to find the angle between two vectors in 2d? A = 2 i + 2 j B = 2 i + 2 j 3d too difficult? Use scalar products: A B = A x B x + A y B y + A z B z = A B cosφ cosφ = A xb x + A y B y + A z B z A B Example 1.11 A = 2 i + 3 j + k B = 4 i + 2 j k

7 How can it be such a great workout with no work? When positive and negative work cancel, the net work is zero even though muscles are exercising.

8 Q6.1 An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. Which statement is correct? A. The cable does positive work on the elevator, and the elevator does positive work on the cable. Cable Motor v Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable.

9 Q6.2 An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct? A. The cable does positive work on the elevator, and the elevator does positive work on the cable. Cable Motor v Elevator B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable.

10 Work done by multiple forces W = ( F ) s tot ( F s ) Work done by the net force Sum of the work done by each force W tot =?

11 We can compare the kinetic energy of different bodies Changes in the energy of a moving body under the influence of an applied force change differently depending on the direction of application. K 1 K 2 What is K 1 :K 2 :K 3? K 3

12 We can compare the kinetic energy of different bodies Changes in the energy of a moving body under the influence of an applied force change differently depending on the direction of application. K 1 : K 2 : K 3 = 1 2 mv 2 : 1 2 (2m)v 2 : 1 2 m(2v)2 =1: 2 : 4 K 1 K 2 Ratio problem like this is your professor s favorite multiple choice question K 3

13 Q6.4 A tractor driving at a constant speed pulls a sled loaded with firewood. There is friction between the sled and the road. The total work done on the sled after it has moved a distance d is A. positive. B. negative. C. zero. D. not enough information given to decide

14 Q6.5 A nonzero net force acts on an object. Which of the following quantities could be constant? A. the object s kinetic energy B. the object s velocity C. both of the above D. none of the above

15 Using Work-Energy theorem, to find speed Example 6.3. From example 6.2: W total = 10kJ Weight=14700N Can be done using acceleration too. Result should be the same! Please perform the exercise using the acceleration to find the speed youself!

16 Application of work-energy theorem A baseball travelling at 20m/s. Mike catches it with his hand, and makes it stop How does these two relate to each other: The distance (s) Mike draw his hand back in order to stop the ball The force (F) Mike receives from the ball? A. The larger s is, the smaller F is B. The larger s is, the larger F is

17 Different objects, different kinetic energies Example: Two boats, same sail Different mass Receive same force Question: Who cross the line first Who has larger kinetic energy at finish line What s the ratio of their speed at finish line?

18 Work and energy with varying forces Figure 6.16 Example: driving a car, alternating your attention between the gas and the brake. The effect: positive or negative force of various magnitude along a straight line. W = x 2 x 1 F x dx

19 The stretch of a spring and the force that caused it The force applied to an ideal spring will be proportional to its stretch. The graph of force on the y axis versus stretch on the x axis will yield a slope of k, the spring constant. What s the work done on the spring?

20 Stepping on a scale (example 6.6) Whether you like the result or not, stepping on a scale is an excellent example of applied force and the work being done to compress that spring. Weight=600N; what is the spring constant k and total work done on it, by you, during the compression?

21 Stepping on a scale F spring (Δx = 1.0cm) = k Δx = w k = w Δx = 600N 0.010m = 60000N /m W you _ on _ spring = 1 2 kx kx 2 1 = 1 2 k(( 0.01m)2 0) = 3.0J

22 Motion with a varying force Glider has intial speed of v 1 =1.50m/s m=0.100kg, k=20.0n/m What is the maximum distance d max? Assuming µ=0 What is the maximum distance d max? Assuming µ=0.47

23 Motion with a varying force Glider has intial speed of v 1 =1.50m/s m=0.100kg, k=20.0n/m What is the maximum distance d max? Assuming µ=0 What is the maximum distance d max? Assuming µ=0.47 W = 1 2 kd 2 = mv 2 W tot = W spring +W f 1 W tot = 1 2 kd 2 max µmgd max = mv 2 1

24 Motion on a curved path Pushing a boy on a swing very slowly to angle θ 0 Boy remain nearly in equlibrium. What is the work done by total force? What is the work done by the tension force What is the work done by the gravity W w = dl = Rdθ W w = φ = 90 o +θ W w = w d l wrcosφdθ θ 0 wr( sinθ)dθ = wr(cosθ 0 1) 0 W F +W w = 0

25 Example: Block B moves down 75.0cm at constant speed How much work is done on B By gravity By the tension in the string By the total force? How much work is done on A By gravity By the tension in the string By friction A By the normal force By the total force B

26 Q6.9 Two blocks are connected as shown. The rope and pulley are of negligible mass. When released, the block of mass m 1 slides down the ramp and the block of mass m 2 moves upward. After each block has moved a distance d, the total work done on m 1 A. is greater than the total work done on m 2. B. is the same as the total work done on m 2. C. is less than the total work done on m 2. D. not enough information given to decide

27 Watt about power? Power: time rate of doing work P av = ΔW Δt Unit: Watt (James Watt) Kilowatt-hour=?J It s work/energy, not Power! Instantaneous power: Power and velocity: P av = F Δs // Δs = F Δt // Δt = F v // av Δt 0 : p = F // v = F v P = dw dt 1 hp=?k W

28 Let s power climb Climbing the 443-m-tall Sears Tower in 15.0 minutes m=50.0kg What s the average power p av = W = mgh t or = (50.0)(9.8)(443) = 241W output by her? p av = F // v av = (mg) h t = 241W

29 Summary Work (Joul, J) done by a force Work done by multiple forces Kinetic energy: K = 1 2 mv 2 W = F s F W tot = ( F ) s s cosφ ( F s ) Work-energy theorem: Work done by varying force: Work done on a curved path: Power (Watt, W): P av = ΔW Δt P = dw dt W tot = K 2 K 1 = ΔK W = W = p = F v x 2 x 1 P 2 P 1 F x dx F d l

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