F) The work by F 6 is zero. The force is perpendicular to the dispalcement.

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1 Phys 00A Homework 6 Chapter 7 All Work and No Play A) The work by is zero. The orce is perpedicular to the dispalcement. B) The work by is positive. The projection o the orce on the direction o the dispalcement is positive. C) The work by is negative. The orce is oposite to the dispalcement. D) The work by is postive. The orce is parallel to the dispalcement. E) The work by 5 is negative. The projection o the orce on the direction o the dispalcement is negative. ) The work by 6 is zero. The orce is perpendicular to the dispalcement. G) The work by 7 is postive. The projection o the orce on the direction o the dispalcement is positive. H) W = dcos θ = (8)(60) = 880 =.9 0 J I) W = dcos θ = (0)(60)cos0 = 57 =. 0 J J) W = dcos θ = ()(60)cos80 = 90 =.9 0 J K) W = dcos θ = (5)(60)cos(80 + 0) = (5)(60)cos0 = 89 =.8 0 J Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant orce o N, one at an angle.0 west o north, and the other at an angle.0 east o north, as they pull the tanker a distance 0.60 km toward the north. N d W E Copyright 00 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No 7

2 James S. Walker, Physics, th Edition Both orces produce the same work. 6 9 total = = cos() = (6 0 )(0.6 0 ) cos() = 00 0 J W W d Conceptual 7. A pendulum bob swings rom point II to point III along the circular arc indicated in the igure.. Picture the Problem: A pendulum bob swings rom point II to point III along the circular arc indicated in the igure at right. Strategy: Apply equation 7-, which says that the work done on an object is positive i the orce and the displacement are along the same direction, but zero i the orce is perpendicular to the displacement. Solution:. (a) As the pendulum bob swings rom point II to point III, the orce o gravity points downward and a component o the displacement is upward. Thereore, the work done on the bob by gravity is negative.. (b) As the pendulum bob swings, the orce exerted by the string is radial (toward the pivot point) but the displacement is tangential, perpendicular to the orce. We conclude that the work done on the bob by the string is zero. Insight: The work done by the Earth is positive i the bob swings rom point I to point II because a component o the displacement is downward and the orce is downward. 7.6 Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lit the. kg pumpkin to a height o. m, then carry it 50.0 m(on level ground) to the check-out stand. 6. Picture the Problem: The pumpkin is lited vertically then carried horizontally. Strategy: Multiply the orce by the distance because during the lit the two point along the same direction. Solution:. (a) Apply equation 7- directly: W = mgd. (b) The orce is perpendicular to the displacement so W = 0. ( )( )( ) =. kg 9.8 m/s. m = 8 J Insight: You can still get tired carrying a pumpkin horizontally even though you re doing no work! Note: The orce o gravity points always down. You exert a orce pointing up to oppose the orce o gravity. You exert positive work while gravity exerts negative work. When Push Comes to Shove Two orces, o magnitudes = 70.0 N and = 50.0 N, act in opposite directions on a block, which sits atop a rictionless surace, as shown in the igure. Initially, the center o the block is at position x i = cm. At some later time, the block has moved to the right, and its center is at a new position, x = 6.00 cm. Copyright 00 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No 7

3 d = x x )xˆ ( i James S. Walker, Physics, th Edition The displacement points to the right when the quantity ( x x i ) is positive. d = x x xˆ = xˆ = ( i) (600 ( 500)) 0 m 0.m A) W = d cos0 = (70)(0.) = 7.7 J B) W = d cos80 = (50)(0.) = 5.5 J C) W = W+ W = =. J net D) Δ K = W net =. J ˆx Conceptual 7. Jogger A has a mass m and a speed v, jogger B has a mass (m/) and a speed v, jogger C has a mass m and a speed (v/), and jogger D has a mass m and a speed(v/).. Picture the Problem: our joggers have a variety o masses and speeds. Strategy: Use the deinition o kinetic energy to determine the relative magnitudes o the kinetic energies. Solution:. Calculate the kinetic energies o each jogger. = =. ( )( ) 9 K ( B = m v = mv ). ( )( ) K m v mv A. ( )( ) ( KC = m v = mv ) 5. ( )( ) KD = m v = mv 6. By comparing the magnitudes o the kinetic energies we arrive at the ranking C < A = D < B. Insight: Even with three times the mass o jogger A, jogger C has only three-ourths the kinetic energy because the kinetic energy is proportional to the square o the speed. 7.9 How much work is needed or a 7 kg runner to accelerate rom rest to 7.7 m/s? 9. Picture the Problem: The runner accelerates horizontally and runs in a straight line. Strategy: The work done equals the change in kinetic energy. Solution: ind the change in kinetic energy: ( ) ( ) W = Δ K = mv mv = = = i 7 kg 7.7 m/s 0 00 J. kj Insight: The runner s kinetic energy comes rom the orces his muscles exert on his center o mass over the distance which his center o mass moves. 7.5 A 0. kg pinecone alls 6 m to the ground, where it lands with a speed o m/s. 5. Picture the Problem: The pine cone alls straight down or 6 m under the inluence o gravity. Strategy: The work done by air resistance is the dierence in kinetic energies between the air resistance and no air resistance cases. The work done by gravity is positive when the object is moving down W g = mgh. Solution:. (a) The total work is equal to the work o gravity plus the work o air resistance. The total work is also equal to the change in kinetic energy. The initial kinetic energy is Wtotal = Wg + W =Δ K = K Ki W = K Wg = mv mgh 0.0 kg m/s 9.8 m/s 6 m 0 J = zero. = ( ) ( ) ( )( ). (b) The work done by air resistance equals the average orce o air resistance times the distance the pine cone alls. It is negative because the upward orce is opposite to the downward distance traveled. ( 0 J) W W W = d so that = = = 0.6 N upward d h 6 m Copyright 00 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No 7

4 James S. Walker, Physics, th Edition Insight: Kinetic riction always does negative work because the orce is always opposite to the direction o motion. 7. A. kg block is held against a spring o orce constant.0 0 N/m, compressing it a distance o 0.0 m.. Picture the Problem: The compressed spring pushes the block rom rest horizontally on a rictionless surace. The block slides to the let as indicated in the igure. Strategy: The spring orce points to the let and the displacement is to the let so the work done by the spring a distance x is kx. The work done by the spring equals the kinetic energy gained by the block. Solution:. Apply equations 7-7 and 7-8: W = kx =Δ K = mv. Now solve or : N/m 0.5 m m/s v v x ( ) k = m = =. kg Insight: The work done on the spring in order to compress it becomes stored potential energy. That stored energy becomes the kinetic energy o the block as the spring accelerates it. at: The uel o Migrating Birds Consider a bird that lies at an average speed o 0.7 m/s and releases energy rom its body at reserves at an average rate o.70 W (this rate represents the power consumption o the bird). Assume that the bird consumes g o at to ly over a distance without stopping or eeding. How ar will the bird ly beore eeding again? d b g at = 9. ood Calories ood Calorie = 000 calories mechanical work=,86 J A) g= g (9. Calorie/g) (,86 J/Calorie)= J The power work relation: W W P = t = = =.5 0 s t P.70 = vt = (07)(.5 0 ) = 55 0 m = 55 km db B) Now we invert the problem to end with the mass o carbohydrates: g carbohydrate =. Calories m carb = J ( Calorie/,86 J) ( g carbohydrate/. Calories)= 8.9 g C) irst we can ind the time or the crossing d t = = = 0 hr (600 s/hr)= 7. 0 s v 0.0 And the work W = Pt = (.70)(7. 0 ) =. 0 J Copyright 00 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No 7

5 James S. Walker, Physics, th Edition m at =. 0 J ( Calorie/,86 J) ( g carbohydrate/9. Calories)=. g Work Horses on Erie Canal Two workhorses tow a barge along a straight canal. Each horse exerts a constant orce o magnitude, and the tow ropes make an angle θ with the direction o motion o the horses and the barge. Each horse is traveling at a constant speed v. A)How much work $W$ is done by each horse in a time $t$? The work per unit time is precisely the power. or each horse W dcosθ d P = = = cosθ = cosθ v t t t W = Pt = vtcosθ B) How much power does each horse provide? Answered above 7.6 An ice cube is placed in a microwave oven. Suppose the oven delivers 05 W o power to the ice cube and that it takes,00 J to melt it. How long does it take or the ice cube to melt? 6. Picture the Problem: The microwave oven delivers energy to the ice cube via electromagnetic waves. Strategy: The power required is the energy delivered divided by the time. W 00 J Solution: Solve equation 7-0 or t: t = = = 07 s = 5. min P 05 W Insight: Power can be regarded as the rate o energy transer because work is essentially transerred energy. We ll learn more about melting ice cubes in Chapter 7 and electromagnetic waves in Chapter 5. Copyright 00 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No 7 5

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