* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solutions to Problems: Work and Energy
Survey
Document related concepts
Fictitious force wikipedia , lookup
Specific impulse wikipedia , lookup
Classical mechanics wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Internal energy wikipedia , lookup
Hooke's law wikipedia , lookup
Mass in special relativity wikipedia , lookup
Atomic theory wikipedia , lookup
Hunting oscillation wikipedia , lookup
Center of mass wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Seismometer wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Kinetic energy wikipedia , lookup
Classical central-force problem wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
Solutions to Problems: Work and Energy P162: 37, 49, 58, 67, 76. 37. The force needed to hold a particular spring compressed an amount x from its normal length is given by F = kx + ax3 + bx 4 . How much work must be done to compressed an amount X, starting from x = 0? Solution: According to the definition, we get X W = ∫ Fdx = ∫ 0 X 0 ( kx + ax 3 + bx 4 )dx = 1 2 1 1 kX + aX 4 + bX 5 . 2 4 5 49. A force of 6.0 N is used to accelerate a mass of 1.0 kg from rest for a distance of 12 m. The force is applied along the direction of travel. The 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: The mass undergoes one-dimensional motion, and thus, (a) the work done by the applied force is W1 = FS = 6.0 × 12 J = 72 J . And (b) the work done by friction can be obtained as W2 = − f k S = − μ k mgS = −0.3 × 1.0 × 9.8 × 12 J = −35 J . (c) From work- energy principle, the final kinetic energy is K 2 = K1 + Wnet = ( 72 − 35 ) J = 37 J . 58. We usually neglect the mass of a spring if it is a small compared to the mass attached to the spring. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length L and mass MS. This mass is uniformly distributed along the length of the spring. A mass m is attached to the end of the spring. One end of the spring is fixed and the mass is allowed to vibrate horizontally without friction (see Fig. 7-30). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed v, the midpoint of the spring moves with speed v/2. Show the kinetic energy of the mass plus spring when the mass is moving with velocity v is K = Mv2/2, where M = m + MS/3 is the “effective mass” of the system. Solution: Assume the length of the spring is X in this case the mass is moving with speed v. Choose a segment element dx in the spring, and it defines a mass element dm = ρ dx = MS 1 x dx , and the kinetic energy reads dK x = ( dm ) vx2 , with vx = v . Thus 2 X X K the kinetic energy of the spring is K S = ∫ dK x = 0 M sv2 2X 3 ∫ X 0 x 2 dx = 1 M s v 2 , and the total 6 kinetic energy of the system is Kt = K S + K m = 1 1 1 1 M S v 2 + mv 2 = Mv 2 , M = M S + m, QED. 6 2 2 3 ( ) 67. A force F = 10.0i + 9.0 j + 12.0k kN acts on a small object of mass 100 g. If the displacement of the object is d = ( 5.0i + 4.0 j ) m, find the work done by the force. What is the angle between F and d . Solution: The work done by the force is W = F id = ( 50 + 36 ) kJ = 86 kJ . And the angle reads θ = arccos F id 86 = arccos = arccos 0.746, or θ = 42° . Fd 325 × 41 76. A simple pendulum consists of a small object of mass m (the bob) suspended by a cord of length L (Fig. 7-33) of negligible mass. A force is applied in the horizontal direction, moving the bob very slowly so the acceleration is essentially zero. (Note that the magnitude of the force needed to vary with the angle that the cord makes with the vertical at any moment.) (a) Determine the work done by this force, to move the pendulum from θ = 0 to θ = θ 0 . (b) Determine the work done by the gravitational force on the bob, FG = mg , and the work done by the force that the cord exerts on the bob. Solution: (a) According to conservation law of the energy, the work done by this force should equal to the change of the potential of the bob: W = mg Δh = mgL (1 − cos θ 0 ) . (b) The kinetic energy of the bob is unchanged during the course and the work done by the tension is zero since the displacement is always perpendicular to the displacement. And the work done by the gravitational force is W = −mg Δh = − mgL (1 − cos θ 0 ) . P192: 6, 7, 32, 67, 70, 79. 6. If U = 3x 2 + 2 xy + 4 y 2 z , what is the force? Solution: The force is the negative gradient of the potential energy function, So ⎛ ∂U ∂U ∂U ⎞ 2 F = −∇U = − ⎜ , , ⎟ = − ( 6 x − 2 y, 2 x + 8 yz, 4 y ) ( N ) . . x y z ∂ ∂ ∂ ⎝ ⎠ 7. A particular spring obeys the force law F = ( − kx + ax3 + bx 4 ) i . (a) Is this force conservative? Explain why or why not. (b) If it is conservative, determine the form of the potential energy function. Solution: (a) The force is conservative, since the work done by it is independent of x2 1 1 1 the path: W = Wx = ∫ Fx dx = − kx2 + ax4 + bx5 x1 2 4 5 x2 x1 . (b) The potential energy function 1 1 1 is of the form: U2 −U1 = −Wx = − kx2 + ax4 + bx5 2 4 5 x1 x2 1 1 1 ⇒ U ( x) = kx2 − ax4 − bx5 , 2 4 5 where we take U ( x = 0 ) = 0 . 32. Consider the track shown in Fig. 8-32. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is horizontal span 3.0m long with a coefficient of kinetic friction μk = 0.25 . The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the tract it is observed to compress the spring by 0.20 m. Determine: (a) the velocity of the block at point B; (b) the work done by friction as the block slides from B to C; (c) the velocity of the block at point C; (d) the stiffness constant k for the spring. Solution: (a) From the work-energy principle, the change of kinetic energy of the block is K f − K i = K f = mgh = 1 2 mv f ⇒ v f = 2 gh = 6.3 m/s . 2 (b) The force of kinetic friction is f k = μk mg , and thus the work done by friction reads W = − f k L = − μk mgL = − 7.4 J . (c) According to work-energy principle, we get K f − Ki = W f ⇒ K f = Ki − f k L ⇒ v f = ( Ki − f k L ) 2m = 4.9 m/s . (d) According to the problem, we learn 1 2 kx = K f ⇒ k = 2 K f x 2 = 6.1×102 N/m . 2 67. The position of a 280-g object is given (in meters) by x = 5.0t 3 − 8.0t 2 − 30t , where t is in seconds. Determine the net rate of work done on this object (a) at t = 2.0 s and (b) t = 4.0 s. (c) What is the average net power input during the interval from t = 0 s to t = 2.0 s, and in the interval from t = 2.0 s to t = 4.0 s? Solution: The velocity and acceleration of the object are vx = dx = 15t 2 −16t − 30 ( m/s) , dt d 2x and a x = 2 = 30t − 16 ( m/s 2 ) , and the net rate of work done on this object dt reads P = ma iv = 63t 3 − 100.8t 2 − 90.16t + 67.2 ( J/s ) , so (a) P P t =4.0s = 4.3×103 J/s . (c) Since vx one can get kinetic energy as K1 t =0s t =0s = −30 m/s, vx = 126 J, K2 t =2.0s t =2.0s t = 2.0s = −25 J/s ; (b) = −2 m/s, and vx = 0.56 J, and K3 t =4.0s t =4.0s = 146 m/s , = 2.98×103 J . According to work-energy principle, we have the work done by the net force during the interval from t = 0 s to t = 2.0 s and in the interval from t = 2.0 s to t = 4.0 s read: W1 = K2 − K1 = −125 J, W2 = K3 − K2 = 2.97 ×103 J , therefore, the average power read: P1 = W1 ( t2 − t1 ) = −63 W, P2 = W2 ( t3 − t2 ) = 1.5 × 103 W . 70. The potential energy of the two atoms in a diatomic molecule can be written as U ( r ) = − a / r 6 + b / r12 , where r is the distance between the two atoms and a and b are positive constants. (a) At what values of r is U(r) a minimum? A maximum? (b) At what values of r is U ( r ) = 0 ? (c) Plot U(r) as a function of r from r = 0 to r at a value large enough for all the features in (a) and (b) to show. (d) Describe the motion of one atom with respect to the second atom when E < 0. (e) Let F be the force one atom exerts on the other. For what values of r is F > 0, F < 0, and F = 0? (f) Determine F as a function of r. Solution: (a) From ∂U ∂r r0 = 0, one can get r0 = ( 2b a ) , and 16 ∂2U ∂2r r0 = 6r0−14 ( 26b − 7ar06 ) = 72br0−14 > 0, So at r0 = ( 2b a ) , U is a minimum. For there is only one stationary point, U 16 approaches its maximum as r approaches zero. (b) From U ( r1 ) = − a / r16 + b / r112 = 0 , one can figure out r1 = ( b a ) 16 . (c) Omitted. ∂U = 6a r 7 − 12b / r13 , ar = 6a m1r 7 − 12b / m1r13 , ∂r and E = K +U < 0, the atom must oscillate between the turning points where E = U. ∂U (e) From Fr = − = 6a r 7 − 12b / r13 , we can easily find that ∂r (d) U ( r ) = − a / r 6 + b / r12 ⇒ Fr = − Fr = 0, for r = r0 = 2b ; Fr > 0 for r < r0 , Fr < 0 for r > r0 . a (f) Fr = − ∂U ∂r = 6a r 7 − 12b / r13 . 79. A ball is attached to a horizontal cord of length L whose other end is fixed, Fig. 8-36. (a) If the ball is released, what will be its speed at the lowest point of its path? (b) a ped is located a distance h directly below the point of attachment of the cord. If h = 0.80 L, what will be the speed of the ball when it reaches the top of its circular path about the peg? Solution: (a) With the conservation law of mechanical law in mind, the speed can be obtained as v = 2 K m = 2mgL m = 2 gL . (b) Similar to case (a), one can get v = 2 K m = 2mg ( L − 2 R ) m = 2 g ( 2h − L ) = 1.2 gL .