Last Time: Finish Ch 5, Start Ch6 Today: Finish Ch 6

Transcription

1 Last Time: Finish Ch 5, Start Ch6 Today: Finish Ch 6 Monday Finish Chapter 5 o Circular mo6on o Dynamics of circular mo6on Chapter 6 o Work done by forces o Kine6c energy o Work and KE prelecture Today Chapter 6 o Work done by non- constant forces o Work and springs o Power o Examples You can pick up your exams AFTER class

2 Varying Forces Up 6ll now we have been assuming force is constant. Graphically we can interpret W=Fs F If instead we have a non- constant force and we look at the F vs x curve. Graphically the work done is s6ll the area under the curve. W # r 2 r 1 F " d r This will be true for any force and any curve. F S S

3 Example work with non- constant forces A block of ice with mass M rests on a fric6onless, horizontal surface. A worker then applies a horizontal force F to it. As a result, the block moves along the x- axis such that its posi6on as a func6on of 6me is given by x(t) = αt + βt 4 +c. a) Calculate the velocity of the object when t=t f. b) Calculate the magnitude of F when t=t f. c) Calculate the work done by the force F during the first 3s of the mo6on. Remember: v = d x dt W # r 2 r 1 W tot = K 2 K 1 K = 1 2 mv2 a = d v dt F " d r

4 Clicker QuesCon A tractor driving at a constant speed pulls a sled loaded with firewood. There is fric6on between the sled and the road. The total work done on the sled aaer it has moved a distance d is A. posi6ve. B. nega6ve. C. zero. D. not enough informa6on given to decide T. S6egler 10/03/2013 Texas A&M University

5 Work and springs Springs are elas6c materials, which means if they suffer some deforma6on from a stretching or compressing force when the force is removed it returns to its original posi6on. This means the spring itself experiences a restoring force. This force is propor6onal to how much it has been deformed. (not always the case but good approxima6on here) The spring constant k has units of N/m F spring = k( r r 0 ) This force is not constant F kx f You will have to use an integral to calculate the work done W # r 2 r 1 F " d r x 0 x f

6 Work and springs F spring = k( r r 0 ) x 0 How much work do you do on the spring to extend it from x 0 to x f? x f How much work does the spring do on you?

7 Example work and springs A woman weighing 600N steps on a scale the has a s6ff spring. In equilibrium the spring is compressed 1.0cm under her weight. Find the spring constant and the total work done on the spring during compression.

8 Clicker QuesCon A box ahached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new loca6on. D During the mo6on the spring does work on the box, your hand does work on the box, and the total work done on the box is? a) posi6ve, nega6ve, zero b) nega6ve, posi6ve, zero c) zero, posi6ve, zero d) nega6ve, zero, zero e) I have no clue

9 Power Power is the rate at which work is done, or work done per unit 6me. Average Power P ave = W t The SI unit of power is the Wah: W=J/s Instantaneous Power W P = lim t"0 t = dw dt Power can also be expressed in terms of the force applied to a body. P = lim t"0 W t = lim t"0 P = F v F # s t

10 Example Average power A 50.0 kg marathon runner runs up the stairs to the top of Chicago s 443m tall Willis Tower. To lia herself to the top in 15.0 min what must be her average power output? In Wahs, kilowahs and in horsepower. Remember: r W F " d r 2 # r 1 P ave = W t P = F v

11 Example Power using force A 0.25 hp motor is used to lia a load at the rate of 5.0 cm/s. What is the biggest load it can lia at this constant speed? (1hp = 746W) Remember: r W F " d r 2 # r 1 P ave = W t P = F v

12 Clicker QuesCon An object is ini6ally at rest. A net force (which always points in the same direc6on) is applied to the object so that the power of the net force is constant. As the object gains speed, A. the magnitude of the net force remains constant. B. the magnitude of the net force increases. C. the magnitude of the net force decreases. D. not enough informa6on given to decide

13 Example Work- energy theorem An 8.00kg sled moves in a straight line on a fric6onless horizontal surface. At one point, its speed is 4.00 m/s; 2.50m beyond this point, its speed is 6.00 m/s. Use Use the work- energy theorem to find the force ac6ng on the sled (assume it is constant and in the same direc6on as the sled s mo6on) Remember: r W F " d r 2 # r 1 W tot = K 2 K 1 K = 1 2 mv2 T. S6egler 06/10/

14 Example e.o.c A block of ice with mass 2.00 kg slides 0.750m down an inclined plane that slopes downward at an angle of 36.9 below the horizontal. If the block starts from rest what is its final speed. (ignore fric6on) Remember: r W F " d r 2 # r 1 W = F s = F s (const. force) W tot = K 2 K 1 K = 1 2 mv2 T. S6egler 10/03/2013 Texas A&M University

15 Example Work done by gravity and fric6on The blocks shown are traveling at a constant speed and move 75.0 cm. a) What is the work done on the 12.0N block by gravity and the tension in the string? b) Work done on the 20.0N block by gravity, tension in the string, and fric6on? c) Total work done on each block

16 Example Work done by gravity and fric6on cont. The blocks shown are traveling at a constant speed and move 75.0 cm. a) What is the work done on the 12.0N block by gravity and the tension in the string? b) Work done on the 20.0N block by gravity, tension in the string, and fric6on? c) Total work done on each block

17 Clicker QuesCon An elevator is being li6ed at a constant speed by a steel cable ahached to an electric motor. Which statement is correct? Cable Motor v Elevator A. The cable does posi6ve work on the elevator, and the elevator does posi6ve work on the cable. B. The cable does posi6ve work on the elevator, and the elevator does nega6ve work on the cable. C. The cable does nega6ve work on the elevator, and the elevator does posi6ve work on the cable. D. The cable does nega6ve work on the elevator, and the elevator does nega6ve work on the cable.

18 Example Work and power A pump is required to lia 800 kg of water (~210ga) per min from a well 14.0 m deep and eject it with a speed of 18.0m/s. a) How much work is done per min in liaing the water? b) How much work is done giving the water the kine6c energy when ejected? c) What must be the power output of the pump? Remember: r W F " d r 2 # r 1 W tot = K 2 K 1 K = 1 2 mv2 P ave = W t P = F v