Ch 7 Kinetic Energy and Work. Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43

Transcription

1 Ch 7 Kinetic Energy and Work Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43

2 Technical definition of energy a scalar quantity that is associated with that state of one or more objects The state of an object describes its position and its motion. motion requires energy: flying a plane requires energy (from the fuel) a thrown ball gets its energy from the thrower

3 One definition of energy is the capacity to do work. This chapter covers kinetic energy, which is just one form of energy, and then work. The work-energy theorem relates the two quantities. SI unit of energy is the Joule 1 J = 1 kg m 2 /s 2 another unit of energy is the calorie

4 Kinetic energy energy associated with motion. Anything that moves has kinetic energy. kinetic energy: K = ½ mv 2 Kinetic energy is related to the mass and the velocity. Eample. A 1 kg ball is thrown with v = 20 m/s K = ½ (1kg)(20m/s) 2 = 200 J

5 Work When you apply a force to an object, you can either accelerate or decelerate the object. Since the velocity changes, the kinetic energy changes. Work is the transfer of energy via a force. Symbol for work is W. Easily confused with weight. Work is a scalar. The unit of work is also the Joule.

6 Relation of work and kinetic energy: Let a force act in the -direction on a particle with 1-D motion: F =ma Multiply equation by ½ m: v 2 v 2 0 2a d substituting 1 2 mv 2 F ma 1 2 mv 2 0 ma d Yields: 1 2 mv mv 2 0 F d F d is the work

7 Work W = F d Work depends on displacement If there is no displacement, there is no work. - If you are holding a book still in the air, you are doing no work - If you are sitting still in a chair, gravity is doing no work. - If you lean on a wall and the wall does not move, you are doing no work.

8 Only components of forces parallel to the displacement do work. F sin F d F cos W = (F cos ) d W = F d cos The vertical component does no work.

9 Only the component of the force that is along the direction of the displacement, does any work. W = F d cos Work is the dot product between the force and the displacement. W F In this definition: W = F d cos F and d are magnitudes, are always positive d The work will be positive or negative based on the cos.

10 Work kinetic energy theorem W = K or K f = K i + W Positive work increases the kinetic energy. Speed up the object Negative work decreases the kinetic energy. Slows down the object

11 Work done by gravitational force The gravitational force is never turned off, so whenever an object has a change in elevation, there is work done by the gravitational force.. W g = mg d cos Since F g is always down, whenever the object is lowered, F g does positive work. When the object is raised, F g does negative work.

12 W g = mg d cos Since F g is always down, whenever the object is lowered vertically, d is down and = 0 0. F g does positive work. When the object is raised straight up, d is up and = F g does negative work.

13 When raising or lowering an object you have to consider the work done by the applied force and the work done by gravity. K = W a + W g If the initial and final velocities of the object being moved are equal then: W a = -W g It does not matter if the K f and K i are zero or not, as long as they are equal. You can lift something at constant velocity, or you can pick up a stationary object and then hold it still.

14 If you have a non-constant force that picked up an object, calculating your work directly may be tricky. However you can calculate the work done by gravity and then take the opposite of W g. Eample of Olympic snatch. The lift requires two separate pulls requiring different muscle groups. Instead of trying to determine how much force each pull eerted and over what distance, you can instead find the work done by gravity and then take: W a = -W g.

15 Eample You lower a bucket down a well, with constant velocity a distance of 8 meters. You eert a constant force of 19.6 N. The bucket has a mass of 2 kg. Find the work done by you, and the work done by gravity. W = mg W = (2kg) g W = 19.6 N 19.6 N W

16 The displacement is down the well. W = (F cos ) Your work: Work by gravity: W y = (19.6 N cos 180) 8m W y = J W g = (19.6 N cos 0) 8m W g = J Notice that the work done by gravity was mgh, where h is the change in height. The net work is zero, the bucket is lowered at a constant velocity.

17 Eample A 500 N force directed 30 degrees above the horizontal is used to pull a 50 kg sled across the ground. The coefficient of friction between the sled and the ground is 0.3. The sled is pulled 5 meters. a) What is the work done by the pulling force b) work done by friction. c) work done by the normal force d) work done by gravity

18 P = 500 N, = 30 0 m = 50 kg = 0.3 = 5 m F f W = mg = 490 N F N W F N + P y = W F N = W P y = mg 500 N (sin 30) F N = (50 kg)g 250 N = 240 N P F f = F N = 0.3*240 N = 72 N

19 W = (F cos ) Work by pulling force: W p = (500 N cos 30) 5 m = 2165 J Work by friction: W f = (72 N cos 180) 5 m = J Work by normal force: W N = (240 N cos 90) 5 m = 0 J Work by gravity: W g = (490 N cos 270) 5 m = 0 J Net work is then 2165 J 360 J = 1805

20 Work by a general variable force When the force is variable we cannot use: W = (F cos ). This is because the work done over each interval, is different. Instead we have to integrate the force over the total displacement to find the work. W i f F( ) d The area under the Force vs. displacement cure is equal to the work. (see fig. 7-13)

21 Let: and F dr F iˆ diˆ Variable force in 3-D F y ˆj dyj ˆ Fzkˆ dzkˆ (the small change in position) The amount of work produced by the force over the small displacement interval dr is: dw F dr F d F dy F dz The work over the total displacement is: rf rf W dw F dr W r i i f F d y r i y y i f z Fydy z z i f Fzdz

22 Work-Kinetic Energy Theorem with a variable force The work of a variable force is: W i f F( ) d Using Newton s 2 nd law we rewrite as: ma d = dv m d dt W i f ( ma) d Using the chain rule we have: W W i f dv dt dv d d dt dv d v f 2 2 ( ma) d m vdv mv f mvi v K i 1 2 v 1 2

23 Spring A spring force is a particular type of variable force. Spring forces are important because many forces behave mathematically like spring forces. Spring can be used to apply forces Springs can store energy These can be done by either compression, stretching, or torsion.

24 Springs Ideal, or linear springs follow a rule called: Hooke s Law: Fs = - k Also an ideal spring will have negligible mass. k is called the spring constant. This determines how stiff the spring is. is the distance the spring is deformed (stretched or compressed) from the equilibrium length. The minus sign tells us that this is a restoring force.

25 Restoring force means that the force the spring eerts, is in the opposite direction of the force that deforms the spring. If I pull the bo to the right (stretch the spring), the spring will eert a force to the left. If I push the bo to the left (compress the spring), the spring will eert a force to the right.

26 Spring constant The spring constant, (k), determines how stiff a spring is. - High spring constant Stiff or strong, Hard to stretch or compress - Low spring constant Limp or weak, Easy to stretch or compress - Units for spring constant are force per length: N/m

27 Simple spring eample A spring with a spring constant of 250 N/m has a length of 0.5 meters when un stretched. What magnitude of force is needed to stretch the spring so that is 0.75 meters long? F s = - k F s = (250 N/m)(0.75m 0.5m) = 62.5 N I got rid of the minus sign to show the magnitude. Note that it would take the same amount of force to compress the spring by 0.25 meters.

28 W = (F cos ) assumes a constant amount of force. So this will not work for a spring. The force needed to deform (stretch/compress) a spring increases as the deformation increases. W W W s s s i i f f 1 2 F d kd k ( k)( f i ) ki kf i f d If i =0, the work of a spring force, where is the deformation of the spring: W s = - ½ k 2

29 When a spring is deformed: The work done by the spring will be positive if the spring ends up closer to its relaed length. W s will be negative if the spring ends up further away from its relaed length. W s is zero if the difference of the spring s length to the relaed length is unchanged.

30 Similar to the case of an applied force acting against gravity. We can find the work of an applied force acting against a spring using: K = W a + W s If the ends of the spring are stationary before and after the displacement. then: W a = -W s

31 Power Definition of power is the work done per time. W P ave t SI unit of power is the watt (W). 1 watt = 1 J/s (Don t confuse the W for watt with the W for work) Instantaneous power: P dw dt

32 U.S. unit of power is horsepower 1 hp = 746 W For electric power generation/usage, use the kilowatt-hour. This is the energy transferred in 1hr at the rate of 1kW (1000 J/s). 1kWh = (1000J/s)(3600s) = J

33 General form for power: P P P dw dt Fv cos F v F(cos dt ) d F cos d dt Power is the dot product of the force and velocity vectors. Only components of forces that are parallel to the velocity do work. The power from forces that are perpendicular to the motion is zero.

34 Power Power is related to how fast a force can be applied. Lifting a heavy weight slowly may not require much power. Picking up the same weight quickly will require more power. Weightlifting eamples: compare the power required to perform a bench press and an Olympic snatch.

35 Assume the weights are moved at constant velocities and the applied forces are constant. Bench press: 300 lbs (1335 N), P = F v Range of motion ~ 0.5 m time to raise weight ~3 seconds = F / t = (1335 N)(0.5 m)/(3s) = W The above assumptions lead us to the power being constant, (equal to the average power).

36 Olympic snatch: 100 lbs (445 N) P = F v Range of motion ~ 1.5 m time to raise weight ~1 s = F / t = (445 N)(1.5 m)/(1s) = W Compared to the power in the bench press (222.5 W) Even though the snatch is performed with less weight, it requires more power because of the larger velocity. (Could have found the power by finding the work of the applied force, using W a = - W g. Then using power = work/time.)

37 Bucket Eample You want to lift a 20 kg bucket up a well at a constant velocity of 0.5 m/s. What power is needed to do so? Since the velocity is constant, the upward force you must pull with is equal to the weight of the bucket. P = F v = F v = (20kg)g (0.5 m/s) = 98 W

38 Another bucket eample Again you want to raise the same 20 kg bucket. It is starting from rest, and you want to pull on the bucket so that is has a velocity of 2 m/s. You want to accomplish this over a time interval of 4 seconds. Use Power and the work energy theorem. (net work equals change in kinetic energy.)

39 first find the change in the kinetic energy. K = ½ mv f2 ½ mv 02 = ½ (20kg)(2m/s) 2 K = 40 J Work = 40 J Average power = Work/time = (40 J) /(4s) = 10 W

40 Shamu eample Calculate the average power needed for the whale to speed up. The killer whale has mass of 8000 kg What power is needed to reach speed of 12m/s in a 6 second time interval? Do work energy theorem K = ½ mv f2 ½ mv 02 = ½ (8000kg)(12m/s) 2 K = J Power = ( J)/6 s = W This was neglecting drag. The power is actually higher W (1hp/746W) = 128 hp About the same as a car.

41 Power delivered by elevator motor A 1000 kg elevator carries a load of 800 kg. A constant friction force of 4000 N retards it upward motion. The retarding force behaves similar to friction. What minimum power in kilowatts and horsepower must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?

42 Elevator Since the speed is constant (3m/s), the acceleration and the sum of all forces equals zero. Find the force (T) the motor must pull with to achieve this motion. F = ma = 0 T F r mg = 0 T = F r + mg = 4000 N + (1800kg)g T = N mg F r T

43 Elevator Now that we know the force the motor must eert we can find the power using P = F v =F v. P = F v = ( N)(3m/s) = W 746 W = 1 hp So P = ( W) (1 hp)/(746 W) = 87 hp

44 Question 8 Problems: 10, 12, 20, 24, 30, 34, 48