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Chapter 7 Units of Chapter 7 Work and Energy Work Done by a Constant Force Scalar Product of Two Vectors Work Done by a Varying Force Kinetic Energy and the Work-Energy Principle 1 2 7-1 Work Done by a Constant Force 7-1 Work Done by a Constant Force The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement: As long as this person does not lift or lower the bag of groceries, In the SI system, the units of work are joules: he is doing no work on it. The force he exerts has no component in the direction of motion. 3 4 7-1 Work Done by a Constant Force 7-1 Work Done by a Constant Force Example 7-2, page 166. (a) Determine the work a hiker must do on Solving work problems: a 15.0-kg backpack to carry it up a hill of 1. Draw a free-body diagram. height h = 10.0 m, as shown. Determine also (b) the work done by gravity on the 2. Choose a coordinate system. backpack, and (c) the net work done on 3. Apply Newton’s laws to determine any unknown the backpack. For simplicity, assume the forces. motion is smooth and at constant velocity (i.e., acceleration is zero). 4. Find the work done by a specific force. 5. To find the net work, either SOLUTION: a) find the net force and then find the work it does, or 1. Draw a FBD and show forces acting on the backpack. b) find the work done by each force and add. 2. Choose a coordinate system. 5 6 7-1 Work Done by a Constant Force 7-2 Scalar Product of Two Vectors 3. Apply Newton’s laws. ∑Fy = may → FH – mg = 0 Definition of the scalar, or dot, product: Hence FH = mg = (15.0 kg)(9.8 m/s2) = 147 N. 4. Work done by a specific force. (a) Work done by the hiker on the backpack is WH = FH (d cosθ) = FH h = mgh = (147 N)(10.0 m) = 1470 J. (b) The work done by the gravity on the backpack is WG = FG d cos(180° - θ) WG = FG d(- cosθ) = mg(-d cosθ) = - mgh = - (15.0 kg)(9.8 m/s2)(10.0 m) = - 1470 J. Therefore, we can write: 5. Net work done. (c) The net work done on the pack is Wnet = WG + WH = - 1470 J + 1470 J = 0. Note: Even though the net work done by all the forces on the backpack is zero, the hiker does do work on the backpack equal to 1470 J. 7 8 7-2 Scalar Product of Two Vectors 7-2 Scalar Product of Two Vectors A·B=B·A Commutative Example 7-4: The force shown has magnitude FP = 20 N and makes an angle of 30° to the ground. Calculate the work done by this force, A · (B + C) = A · B + A · C Distributive using the dot product, when the wagon is dragged 100 m along the ground. Unit vectors i·i= j·j=k·k=1 i·j=i·k=j·k=0 If A = Ax i + Ay j + Az k B = Bx i + By j + Bz k then A · B = AxBx + AyBy + AzBz Solution: W = FP · d FP = (FP cos 30°) i + (FP sin 30°) j = 17 N i + 10 N j and d = (100 m) i Special case A = B W = FP · d = (17 N)(100 m) + (10 N)(0) = 1700 J. A · A = Ax 2 + Ay 2 + Az 2 = A2 . 9 10 7-3 Work Done by a Varying Force 7-3 Work Done by a Varying Force Particle acted on by a varying force. Clearly, F · d is not constant! For a force that varies, the work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up. The work done ΔWi by the force Fi for the i-th interval is Δwi ≈ Fi cosθi Δℓi 11 12 7-3 Work Done by a Varying Force 7-3 Work Done by a Varying Force In the limit that the pieces become infinitesimally narrow, the Work done by a spring force: work is the area under the curve: The force exerted by a spring is given by Hooke’s law: or: . k is called the spring constant. Note the negative sign… Force exerted by the spring is always directed opposite the displacement. 13 14 7-3 Work Done by a Varying Force 7-4 Kinetic Energy and the Work-Energy Principle If we write the acceleration in terms of the velocity and the distance, Plot of F vs. x. we find that the work done is Work done is equal to the ⎛ v2 − v12 ⎞ 2 ⎛ v2 − v12 ⎞ 2 shaded area. Wnet ⎜ 2d ⎟ d = m⎜ 2 ⎟ = Fnet d = mad = m⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ or We define the quantity mv2/2 to be the translational kinetic energy K of the object: 15 16 7-4 Kinetic Energy and the Work-Energy Principle Problem 49 (3rd ed.) A force of 6.0 N is used to accelerate a mass of 1.0 kg from rest for a This means that the work done is equal to the change in the distance of 12 m. The force is applied along the direction of travel. The kinetic energy: coefficient of kinetic friction is 0.30. What is the work done (a) by the applied force? (b) by friction? (c) What is the kinetic energy at the 12-m mark? Solution: (a) The work done by the applied force is WF = F d = (6.0 N)(12 m) = 72 J • If the net work is positive, the kinetic energy increases. (b) The work by friction is • If the net work is negative, the kinetic energy decreases. Wfr = FN d cos 180° = μk mg d (-1) = - (0.30)(1.0 kg)(9.8 m/s2)(12 m) = -35 J Because work and kinetic energy can be equated, they must (c) The normal force and weight do no work. The net work increases the kinetic have the same units: kinetic energy is measured in joules. energy of the mass: Wnet = WF + Wfr = ΔK 72 J – 35 J = Kf – 0 which gives Kf = 37 J. 17 18 Problems 47 (3rd ed.) Problem 11 At an accident scene on a level road, investigators measured a car’s A 380-kg piano slides skid mark to be 78 m long. It was a rainy day and the coefficient of 3.9m down a 27° kinetic friction was estimated to be 0.38. Use these data to determine incline and is kept from the speed of the car when the driver slammed on (and locked) the accelerating by a man brakes? who is pushing back on it parallel to the Solution: incline. Determine: (a) On a level road, the normal force is mg, so the kinetic friction force is the force exerted by μkmg. We have the man, (b) the work done by the man on W = ΔK → μkmg d cos 180° = ½ mv2 - ½ mv0 2 the piano, (c) the work done by the force of (0.38) m (9.8) (78) (-1) = - ½ mv0 2 gravity, and (d) the net work done on the v0 = 24 m/s (87 km/h) piano. Ignore friction. 19 20 End of Problem 11 # 11 Chapter Problems chapter problems End ofProblem 11 # 11 r . FN (b) WP = FPd cos180° = − (1691N)( 3.9m) = −6595J ≈ −6600J r (a) FP y θ ( ) (c) W = F d cos63° = mgd cos63° = ( 380kg) 9.80m s2 ( 3.9m) cos63° r mg x G G = 6594N ≈ 6600J ∑F y = FN − mg cos θ = 0 → FN = mg cos θ ∑F x = mg sin θ − FP = 0 → (d) Since the piano is not accelerating, the net force on the piano is 0, and so the net work done on the piano is also 0. This FP = mg sin θ = mg sin θ can also be seen by adding the two work amounts calculated. ( = ( 380 kg ) 9.80 m s2 ) ( sin 27°) = 1691N ≈ 1700 N Wnet = WP + WG = −6.6 × 103 J + 6.6 × 103 J = 0 J 21 22 End ofProblem 13 # 13 chapter problems End ofProblem 13 # 13 chapter problems A 17,000-kg jet takes off from Solution: an aircraft carrier via a catapult. The gasses thrust out (a) The gases exert a force on the jet in the same direction as the of jet’s engines exert an displacement of the jet. From the graph we see the displacement constant force of 130 kN on the of the jet during launch is 85 m. jet; the force exerted on the jet by the catapult is plotted in ( ) Wgas = Fgas d cos 0° = 130 × 103 N ( 85m ) = 1.1 × 107 J figure (b). Determine: (a)The work done on the jet by (b) The work done by catapult is the area underneath the graph in the gases expelled by its Figure 7-22. That area is a trapezoid. engines during the launch of the jet; and (b)The work done on the jet by Wcatapult = 1 2 (1100 × 10 N + 65 × 10 N ) (85m) = 5.0 × 10 J 3 3 7 the catapult during launch of the jet. 23 24 End of Problem 54 # 54 chapter problems Chapter Problems End of Problem 54 # 54 First find the kinetic energy of the train, and then find out how Spiderman uses his much work the web must do to stop the train. spider webs to save a runaway train, Note that the web does negative work, since the force is in the moving at 20 m/s. His opposite direction of the displacement. web stretches 800 m before the 104 kg train comes to a stop. Assuming the web Wto stop = ΔK = 1 mv2 − 1 mv12 = 0 − 1 104 kg ( 20 m s ) = −2 × 106 J train 2 2 2 2 ( ) 2 acts like a spring, estimate the spring constant. Wweb = − 1 kx 2 = −2 × 10 6 J → k = ( )= 2 2 × 10 6 J 6N m 2 (800 m ) 2 25 26 End of Problem 63 # 63 Chapter Problems End of Problem 63 # 63 Chapter Problems (a) The angle between the pushing force and the displacement is 32°. (a) How much work is done by the horizontal force FP = 150 N on the 18-kg block when the force pushes the block 5.0 m up along the 32° WP = FPd cos θ = (150 N )( 5.0 m) cos32° = 636.0J ≈ 640J frictionless incline? (b) How much work is done by the gravitational force on the block during this displacement? (c) How much work is (b) The angle between the force of gravity and the displacement is 122°. done by the normal force? (d) What is the speed of the block (assume that it is zero initially) after this displacement? ( ) WG = FGd cosθ = mgd cosθ = (18kg) 9.80m s2 ( 5.0m) cos122° = −467.4J ≈ −470J (c) Because the normal force is perpendicular to the displacement, the work done by the normal force is 0. (d) The net work done is the change in kinetic energy. W = WP + Wg + WN = ΔK = 1 2 mv2 − f 1 2 m v i2 → 2W 2 ( 636.0 J − 467.4 J ) vf = = = 4.3 m s 27 m (18 kg ) 28

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