Work and Direction. Work and Direction. Work and Direction. Work and Direction

Transcription

1 Calculate the net gravitational force on the shaded ball. Be sure to include the magnitude and direction. Each ball has a mass of 20,000 kg. (0.79N, 22.5 o N of E) Chapter Six Work = Force X distance W = Fd Unit - Joules (N m), (kg m 2 /s 2 ) Force must be in the direction of the motion. Work and Direction Work and Direction Lifting a box is work Fighting force of gravity. Lifting force in same direction box moves. Your lifting force and the box s direction Pushing a box is work Applied force is in the direction of movement Working against force of friction Pushing on ice would be less work Frictional Force Fighting against the force of gravity Work and Direction Work and Direction Carrying a box is not work But this is the direction of motion Your lifting force Fighting against the force of gravity Mowing the lawn is some work only do work in horizontal direction. The downward push, because of the angle of the handle, doesn t do any work. This part is lost Your push This is the only part that does work 1

2 Vertical vs. Horizontal Work Lifting Gives an object Potential Energy of position. Pushing Gives an object Kinetic Energy of Motion. How much work is this man doing carrying the cat. Assume the cat masses 8 kg. Work or Not 1. A teacher pushes against a wall until he is exhausted. 2. A book falls off the table and falls freely to the ground. 3. A waiter carried a full tray of meals across the room. 4. A rocket accelerates through space. 5. A rocket travels at a constant speed through space How much work is done lifting a 5.0 kg gnome to a shelf 2.0 meters above the floor at a constant speed? Superman does 36,750 J of work lifting this car 2.5m from the ground. What is the weight of the car? What is the mass of the car? A waiter lifts a 6.00 kg platter 35.0 cm from the kitchen. a. Calculate the work done lifting the platter. (20.6 J) b. Calculate the work done carrying the platter to the table. c. Suppose they had put the platter on a cart and pushed it to the table. They used a force of 25.0 N for 10.0 m. Calculate the work. (250 J) d. What were they most likely working against? 2

3 Mr. Fredericks pulls a box with 30 N of Force a distance of 5.0 m, at an angle of 50 o with the ground. a. Calculate the work was done (96.5 J) b. If the box masses 45.0 kg, calculate the normal force (418 N) F x = (30N)(cos50 o ) = 19 N F y = (30N)(sin50 o ) = 23 N W = F x d = (19N)(5m) = 96J W = Fdcosq Direction of motion F y q = 50 o q = 50 o F x A sled is pulled with a force of N at an angle of 65 o to the ground for 8.00 m. a. Calculate the work done. (338 J) b. Calculate the normal force (the sled masses 20.0 kg) (105 N) A 50-kg crate is pulled 40 m with a force of 100 N at an angle of 37 o. The floor is rough and exerts a frictional force of 50 N. Determine the work done on the crate by each force and the net work done on the crate. F p f k n mg q Let s deal with the vertical forces first W G = mgxcos90 o = 0 W N = F N xcos90 o = 0 Now the horizontal forces W fr = F fr d W fr = (50 N)(40m) = J No work is done in the vertical direction (the box is not lifted) W p = F p dcos 37 o W p = (100N)(40 m)(cos 37 o ) = 3200 J W net = W p + W fr W net = 3200 J J = 1200 J 3

4 A 35.0 kg suitcase is pulled 20.0 m with a force of N at an angle of 45.0 o. The floor has a m k of a. Calculate the normal force on the suitcase (HINT: take the F py into account). (272 N) b. Calculate the work done by friction. (-1088 J) c. Calculate the work done by the pull. (1414 J) d. Calculate the net work on the suitcase. (326 J) A hiker walks up a hill at a constant speed. He is carrying a 15.0 kg backpack and the hill is 10.0 m high. How much work did he do? No work in the horizontal direction. f k n mg F p q The only forces are vertical 0 =F v -mg F v = mg = (15.0 kg)(9.80 m/s 2 ) F v = 147 N W hiker = Fd = (147 N)(10.0 m) W hiker = 1470 J How much work did gravity do? W G = J W NET = W hiker + W G W NET = 1470 J J = 0 Does the Earth Do Work on the Moon? W = Fdcosq W = Fd(cos 90 o ) W = Fd(0) W = 0 v F R English Unit of Work Foot-pound English unit of work. Pound unit of Force W = Fd = (foot*pound) 4

5 Variable Force Work is really an area: Given the following graph, determine the work done by the following force in the first ten meters W = Fdx (an integral tells you the area) WORK Calculate the work: a) In the first 10 meters b) In the second 10 meters c) Overall Calculate the work: a) In the first 20 meters b) From 20 to 40 meters c) From 40 to 80 meters d) Overall Calculate the work: a) In the first 10 meters b) In the second 10 meters c) Overall Calculate the work: a) In the first 10 meters b) In the second 10 meters c) Overall 5

6 A 50.0 kg child and sled is pulled by a rope with a tension of N and an angle of 35.0 o with the ground for 50.0 m. The sled and snow has a m k of a. Calculate the normal force on the sled. (375 N) b. Calculate the work done by friction. ( J) c. Calculate the work done by the pull. (8191 J) d. Calculate the net work on the sled. (5378 J) A 70.0 kg student is pulled by Mr. Fredericks with a force of N and an angle of 40.0 o with the ground for 75.0 m to the office. The floor has a m k of a. Calculate the normal force on the student. (365 N) b. Calculate the work done by friction. ( J) c. Calculate the work done by the pull. (28,725 J) d. Calculate the net work. (21,900 J) Energy Energy The capacity to do work 1 Joule = 1 Newton meter =1 kgm 2 /s 2 Two types Potential Energy Kinetic Energy Kinetic Energy Definition - Energy of motion K = ½ mv 2 m = mass (kg) v = speed (m/s) Unit = Joules Questions: a. If the mass of an object is doubled, what happens to KE? b. If the speed is doubled, what happens to KE? A cat is traveling with a kinetic energy of 10.0 J at 2.00 m/s a. Calculate his mass b. If Mr. Fredericks (56.0 kg) has the same kinetic energy, calculate his speed. Deriving Kinetic Energy W = Fd W = mad W = m(v f 2 v i2 )d 2d W = ½ mv f 2 ½ mv i 2 W NET = K f K i W NET = DK #only if no change in elevation 6

7 + (Fd) General Form W = DK + DU How much work is required to accelerate a 1000-kg car from 20-m/s to 30-m/s? - (f k d) ALWAYS MAKE FRICTION NEGATIVE (W = 250,000 J) A 1000 kg car stops over 50.0 m. The coefficient of friction is 0.25 a) Calculate the force of friction. (-2450 N) b) Calculate the work done by friction. ( X 10 5 J) c) Calculate the initial speed of the car. (15.7 m/s) A g football travelling at 25.0 m/s is caught by a player. The players arms go back 75.0 cm while catching the ball. a) Calculate the initial kinetic energy of the ball. (125 J) b) Calculate the force exerted on the ball by the player s hands. (-167 N) A g baseball is caught by a fielder. The glove moves 25.0 cm. The fielder experienced an average force of 204 N. a. Calculate the kinetic energy of the ball before being caught. (51 J) b. Calculate the initial speed of the ball. (27.0 m/s) c. Draw a free body diagram of the ball in the air. d. Draw a free body diagram of the ball while being caught by the glove. A 2.50 kg box is given an initial speed of 2.00 m/s. If the m k between the box and the ground is 0.300, calculate how far the box will slide. 7

8 Potential Energy Definition - Stored energy Two Types Energy of position Gravitational PE - stored by placing something at a height above the ground Mechanical PE - compressing a spring or mechanical device Energy stored in chemical bonds and atomic nuclei Potential Energy of Position U = mgy m = mass in kg g = acceleration of gravity (9.8 m/s 2 ) y = height from the ground in meters. Potential Energy of Position. What would the potential energy of a ball be at the other steps? 45 J 15 J Potential Energy and Work How much work does a 50.0 kg woman do by climbing to the top of a 300 m hill? W = DK + DU W = DU #only if no change in speed W = mgy f mgy i 8

9 Assume the roller coaster car has a mass of 1000 kg. Calculate the PE: a) At pts. A, B, and C b) Gained from A B c) Lost B A d) Lost B C F x = kx Hooke s Law F = Force exerted on the spring k = spring constant x = distance compressed or stretched. A force of 600 Newtons will compress a spring 0.5 meters. a) Calculate the spring constant of the spring. (1200 N/m) b) Calculate the force necessary to stretch the spring by 2 meters. (2400 N) c) A force of 40 Newtons will stretch a different spring 0.1 meter. How far will a force of 80 Newtons stretch it? (0.2 m) U = ½ kx 2 Springs k = spring constant (measure of the stiffness of a spring) Spring Example 1 How much potential energy is stored in a spring if it is compressed 10.0 cm from its normal length. Assume it has a spring constant of 300 N/m x = distance stretched from normal length U = ½ kx 2 U = ½ (300 N/m)(0.100 m) 2 U = 1.5 N m = 1.5 J 9

10 When 10 J is used to stretch a spring, the spring stretches 20.0 cm. Calculate the spring constant. 50 N of force is required to stretch a spring by 0.50 m. Calculate the spring constant. Calculate how much potential energy is stored in the spring. Conservative and Nonconservative Forces Conservative Forces Work is independent of the path taken Gravity, elastic spring Nonconservative Forces Work depends on the path taken Friction, air resistance, tension in a cord Also called dissipative forces Conservative Forces 6 floors 4 floors Imagine carrying a box up 6 floors, then down 2. Overall, you only increased the PE by 4 floors (gravity is conservative). What if you went up 8 and down 4? Nonconservative Forces Will it take more work to push the box from 1 to 2 on path A or path B? Or are they the same? Law of Conservation of Mechanical Energy W = DK + DU (assume no friction or outside force) 0 = K f - K i + U f - U i B K i + U i = K f + U f ½ mv 2 i + mgy i = ½ mv 2 f + mgy f A 10

11 Mr. Fredericks (56.0 kg) climbs up a 2.50 m slide. Calculate the speed at which he will leave the slide at the bottom. Neglect friction. Suppose Mr. Fredericks wanted to leave the slide with a speed of 10.0 m/s. Calculate the height of the slide without using the mass. Neglect friction. Law of Cons. Energy: Ex 1 Mr. Fredericks (100 kg) jumps from a 3 meter cliff. Calculate his speed when he is: a) 2 m above the ground b) 1 m above the ground c) Just before he hits (0 m) a) 2 m above the ground b) 1 m above the ground ½ mv mgy 1 = ½ mv mgy mgy 1 = ½ mv mgy 2 mgy 1 = m(½v gy 2 ) gy 1 = (½v gy 2 ) (9.8 m/s 2 )(3m) = ½v (9.8 m/s 2 )(2m) v 2 2 = 19.6 m 2 /s 2 v 2 = 4.4 m/s ½ mv mgy 1 = ½ mv mgy mgy 1 = ½ mv mgy 2 mgy 1 = m(½v gy 2 ) gy 1 = (½v gy 2 ) (9.8 m/s 2 )(3m) = ½v (9.8 m/s 2 )(1m) v 2 2 = 39.4 m 2 /s 2 v 2 = 6.3 m/s 11

12 c) 0 m above the ground (just before he hits) ½ mv mgy 1 = ½ mv mgy mgy 1 = ½ mv mgy 1 = ½ mv 2 2 (All PE gets turned to KE) gy 1 = ½v 2 2 v 2 2 = (9.8 m/s 2 )(3 m)(2) v 2 = 7.7 m/s Law of Cons. Energy: Ex 2 A rollercoaster starts at a height of 40 m. a) What is its speed at the bottom of the hill? (28.0 m/s) b) At what height will it have half of that speed? (30.0 m) 40 m a) What is its speed at the bottom of the hill? b) At what height will it have half of that speed? ½ mv mgy 1 = ½ mv mgy mgy 1 = ½ mv gy 1 = ½ v 2 2 v 2 = \/ 2gy 1 = \/ (2)(9.8 m/s 2 )(40 m) v 2 = 28 m/s ½ mv mgy 1 = ½ mv mgy mgy 1 = ½ mv mgy 2 gy 1 = ½ v gy 2 (9.8 m/s 2 )(40 m) = ½ (14 m/s) 2 + (9.8 m/s 2 )y 2 y 2 = 30 m (above the low point) A narwhal on a water slide is at a height of 25.0 m, and assume it starts out with almost no speed at the top. a) Calculate the speed at the bottom. (22.1 m/s) b) Calculate the height at which it will have 75% of that speed. (10.9 m) Law of Cons. Energy: Ex 3 Superman is supposed to be able to leap tall buildings in a single bound. What initial velocity would he need to jump to the top of the Empire State Building (381 m)? 12

13 ½ mv 2 gr + mgy gr = ½ mv 2 top + mgy top ½ mv 2 gr + 0 = 0 + mgy top ½ mv 2 gr = mgy top ½ v 2 gr = gy top v gr = \/2gy top = (2 X 9.8m/s 2 X 381 m) 1/2 v gr = 86.4 m/s (193 mph) What if he fell? v max = mph. Equivalent to jumping from about 150 meters (about 450 feet). Springs and Conservation of Energy Springs store potential energy W = DK + DU + DU sp (work is usually zero) Springs: Example 1 A kg toy dart compresses a spring 6.0 cm. The spring s constant is 250 N/m. What speed will the dart leave the gun? ½ mv ½ kx 2 1 = ½ mv ½ kx ½ kx 2 1 = ½ mv ½ kx 2 1 = ½ mv 2 2 kx 2 1 = mv 2 2 v 2 2 = kx 2 1/m =(250 N/m)(0.060m) 2 /(0.100 kg) v 2 2 = 9.0 m/s 2 v 2 = 3.0 m/s A 1.50 kg block slides along a smooth table and collides with a spring (k = 1000 N/m). If block had an initial speed of 1.20 m/s, how far was the spring compressed? (4.65 cm) 13

14 Springs and Vertical Movement Car springs/struts Bungee Cord Make the zero the unstretched spring A 2.60 kg ball is dropped. It falls 55.0 cm before hitting a spring. It compresses the spring 15.0 cm before coming to rest. Calculate the spring constant. ½ mv mgy 1 + ½ ky 2 1 = ½ mv mgy 2 + ½ ky mgy = 0 + mgy 2 + ½ ky 2 2 mgy 1 = mgy 2 + ½ ky cm (2.60 kg)(9.8m/s 2 )(0.550m) = (2.60 kg)(9.8m/s 2 )(-0.150m)+ ½ k(-0.150m) cm 14.0 J = J + ( m 2 )(k) J = ( m 2 )(k) k = 1580 N/m A vertical spring has a spring constant of 450 N/m and is mounted on the floor. A 0.30 kg block is dropped from rest and compresses the spring by 2.50 cm. Calculate the height from which the block was dropped. Use the unstretched spring length as your zero point. A vertical spring has a spring constant of 895 N/m is compressed by 15.0 cm. a) Calculate the upward speed can it give to a kg ball when released. (7.28 m/s) b) Calculate how high above the top of the uncompressed spring the ball will fly. (2.70 m) (2.28 cm) 14

15 A child s toy shoots 2.70 g ping pong balls. When a ball is loaded into the tube, it compresses the spring (k = 18 N/m) by 9.5 cm. If you shoot a ping pong ball straight up out of this toy, how high will it go? A 300 g block is dropped from rest 40.0 cm above a spring. It compresses the spring by 9.50 cm. Calculate the spring constant [2.97 m] (322 N/m) A rock rolls down a road inclined 12 o to the horizontal. It rolls 2400 m along the road. How fast is the rock going when it reaches the bottom, striking a lazy physics student? A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground. a. Determine the kinetic energy possessed by the skydiver. [150,000 J] b. Determine the potential energy possessed by the skydiver. [665,000 J] c. Determine what height she jumped from. [1066 m] d. Determine the speed when she is m above the ground. [105 m/s] A skier glides down a frictionless hill of 100 meters, then ascends another hill, of height 90 meters. What is the speed of the skier when it reaches the top of the second hill? [14 m/s] 15

16 Law of Conservation of Mechanical Energy If nonconservative forces act, use: W fr = DK + DU (W fr is negative) Mr. Fredericks (100 kg) slides down a 3.5 m tall slide. He leaves the slide at a speed of 6.3 m/s. a) Calculate the Force of friction. (-241 N) b) Calculate the coefficient of friction for the slide.(0.303) c) Would there be more or less friction if you raise the left side of the slide. 3.5 m 6.0 m ½ mv mgy 1 = ½ mv mgy 2 + F fr d 0 + mgy 1 = ½ mv F fr d mgy 1 = ½ mv F fr d F fr d = mgy 1 - ½ mv 2 2 F fr = (mgy 1 - ½ mv 2 2)/d F fr = [(100 X 9.8 X 3.5) (½ X 100 X )]/6.0 F fr = 241 N To calculate the coefficient of friction: F fr = mmgcosq (mgcosq is the Normal Force) tan q = 3.5/6.0 q = 35.7 o m = F fr /mgcosq m = 241 N/[(100 kg X 9.8 m/s 2 )(cos 35.7 o )] m =

17 Friction: Example 2 A 1000 kg roller coaster starts from a height of 40 m. On the second hill, it only rises to a height of 25 m. Calculate the force of friction. Assume a travel distance of 400 m. ½ mv mgy 1 = ½ mv mgy 2 + F fr d 0 + mgy 1 = 0 + mgy 2 + F fr d mgy 1 = mgy 2 + F fr d F fr = (mgy 1 - mgy 2 )/d F fr = (1000 X 9.8 X X 9.8 X 25)/400 F fr =368 N [368 N] A delivery boy wishes to slide a 2.00 kg package up a 3.00 m long ramp. The ramp makes a 20 o angle with the ground, and has a coefficient of friction of a) Calculate the height of the ramp (1.03 m) b) Calculate the normal force on the package (18.4 N) c) Calculate the minimum speed the package must have to go up the ramp. Remember to take Potential Energy and the energy lost to friction into account. (6.5 m/s) d) Draw free-body diagrams for the box at the bottom and in the middle of the ramp. Mr. Fredericks (56.0 kg) slides down a 2.50 m tall slide. The slide has a coefficient of and makes a 45.0 o angle with the ground. a) Calculate the length of the slide. (3.53 m) b) Calculate the normal force. (388 N) c) Calculate the force of friction. (97.0 N) d) Calculate how fast he will leave the slide. (6.06 m/s) 2.50 m Power Definition rate at which work is done A powerful engine can do a lot of work quickly. Running and walking up the steps require the same amount of work. Running up steps requires more Power Power Power = Work P = W time t Metric Unit: Joules/s = Watt. 17

18 A donkey performs 15,000 J of work pulling a wagon for 20 s. What is the donkey s power? An eagle snatches a 5.00 kg rabbit from the ground and flies up 10 m with it in 0.75 seconds, what is the eagle s power? Suppose Mr. Fredericks (56.0 kg) can run up the Empire State Building in 2.00 seconds.the Empire State Building is 380 m tall. a. Calculate the work done (2.09 X 10 5 J) b. Calculate the power (1.05 X 10 5 W) c. How could the power be increased? Example 4 A 60-kg runner and a 100-kg runner run 50 m up a 6.0 o slope. Calculate the work done by both runners. First we need to find the height they rise: sin 6.0 o = y 50 m y = 5.23 m 6.0 o 50 m y Remember that work is a change in energy W = DU W = U f U i W = mgy f 0 W 60 = mgy = (60.0 kg)(9.8m/s 2 )(5.23 m) W 60 = 3075 J W 100 = mgy = (100.0 kg)(9.8m/s 2 )(5.23 m) W 100 = 5125 J If it takes the 60 kg man 6 seconds, and the 100 kg man 10 seconds, who has more power? P = W/t P 60 = 3075 J/6 s = 513 W P 100 = 5125 J/10 s = 513 W 18

19 A 20.0 kg child in a 30.0 kg wagon is pushed from rest. The wagon leaves with a speed of 6.00 m/s, and the person pushing has a power of Watts. a) Calculate the work done on the child and wagon. (900. J) b) Calculate the time of the push. (9 s) c) If the coefficient of kinetic friction is 0.25, calculate how far the wagon will travel. (7.35 m) Horsepower The English Unit of power is horsepower Foot-lb = Horsepower (hp) second 1 hp = 746 Watts 1 hp = ½ Columbus (who sailed in 1492) Horsepower: Example 1 How much horsepower is required to power a 100 Watt lightbulb? Horsepower Consider a 100 hp car engine that can go from 0 to 60 mi/hr in 20 seconds. A 400 hp car could go from zero to 60 mi/hr in 5 seconds. 4 times as powerful means it can do the same work in ¼ the time. Horsepower: Example 2 Suppose you have a 1.34 hp lawnmower. How many Watts of power does it have? Mr. McKeown carries a 200 N box of Halloween costumes 5 meters upstairs in 3 seconds. What is his power in Watts and in horsepower? 19

20 You leave a 60 W lightbulb on outside your house for 12 hours a day for an entire month (30 days). How many kilowatt hours will you be charged for? You decide to turn your computer (75 W) off every night. Assume that it is off for 9 hours every night for 30 days. How many kilowatt-hours did you save? Suppose you are charged for 3000 kwh for a given month. How many Joules of energy did you use that month? (Hint: Start from the power formula) Suppose you are charged for 1500 kwh for a given month. How many Joules of energy did you use that month? (Hint: Start from the power formula, and just use one hour) P = Fv Calculate the drag force if a car has 200 hp and travels at 25.0 m/s. Calculate the power required to travel on a flat road against a drag force of 700 N, at a speed of 22 m/s. 20

21 Calculate the force required to climb a 10.0 o incline. Assume an air drag of 700 N and a speed of 22 m/s. However, include the force of gravity. The car has a mass of kg. Also calculate the power g (3082 N, 67,804 W) W = Fd + (Fd) Formula Wrap-Up W = DK + DU (+ DU sp ) - (f k d) P = W/t P = Fv (moving against air drag) 2. a) 1.1 X 10 3 J b) 5.4 X 10 3 J N J J 10.a) 420 N b) J c) J d) 5900 J e) J 18. Square root 2, X 10 5 J m/s N N J 32. a) 45.3 J b) 12.3 J c) 45.3 J 34. a) ½ k(x 2 -x o2 ) b) ½ kx o 2 (Same) m/s m/s 40. v b = 24 m/s, v c = 9.9 m/s, v d = 19 m/s 42. a) 100 N/m b) 22 m/s J m/s m/s m/s s hp kw m/s 21

22 g(2h L) = v 2 /2 2g(2X0.8 L) = v 2 2g(1.6 L) = v 2 22

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25 The value of the acceleration due to gravity was determined and compared to the accepted value of 9.8 m/s 2. Instantaneous velocities were determined using a photogate timer, and a velocity-time graph was generated. The slope of the graph was the experimental value, 9.4 m/s 2. The lab was accurate with a 4.0 % error. 25