fisica1-youn-e-freedman-exercicios-resolvidos

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2 4.18: a) From Eq. (4.9), m = w/ g = (3.20 N) /(9.80 m / s ) = 0.327 kg. 2 b) w = mg = (14.0 kg)(9.80 m /s ) = 137 N. 4.19: F = ma = (55 kg)(15 m /s 2 ) = 825 N. The net forward force on the sprinter is exerted by the blocks. (The sprinter exerts a backward force on the blocks.) 4.20: a) the earth (gravity) b) 4 N, the book c) no d) 4 N, the earth, the book, up e) 4 N, the hand, the book, down f) second g) third h) no i) no j) yes k) yes l) one (gravity) m) no 4.21: a) When air resistance is not neglected, the net force on the bottle is the weight of the bottle plus the force of air resistance. b) The bottle exerts an upward force on the earth, and a downward force on the air. 4.22: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N. ∑ F 620 N−650 N = = 0.452 m/s 2 . The passenger’s 2 acceleration is 0 .452 m / s , downward. m 2 − 650 N/9.80 m/s 2 F mg ( 45 kg)(9.80 m /s ) −23 2 4.23: a E = = = = 7.4 × 10 m /s . 24 m m (6.0 × 10 kg) E E
4.24: (a) Each crate can be considered a single particle: F AB (the force on m A due to m B ) and F BA (the force on m B due to m A ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them. 4.25: The ball must accelerate eastward with the same acceleration as the train. There must be an eastward component of the tension to provide this acceleration, so the ball hangs at an angle relative to the vertical. The net force on the ball is not zero.
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Física I - Sears, Zemansky, Young
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5 1.1: 1mi × ( 5280ft mi) × ( 12
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1.15: a) If a meter stick can measu
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8 11 1.28: The moon is about 4× 10
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1.34: r o o 1.35: A ; A x = ( 12.0
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1.39: Using components as a check f
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1.43: a) The magnitude of A r + B r
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1.48: a) ˆ ˆ ˆ 2 2 2 i + j + k =
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1.53: Use of the right-hand rule to
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1.59: a) To three significant figur
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1.65: Let D r be the fourth force.
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1.68:a) R = A + B + C x R y = x x x
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1.70: The third leg must have taken
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−1 200−20 o 1.73: a) Angle of f
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1.75: Let +x be east and +y be nort
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1.78: (a) Take the beginning of the
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1.83: a) Parallelog ram area = 2 ×
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1.88: a) This is a statement of the
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1.91: A r and C r are perpendicular
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1.96: a) b) i) In AU, ii) In AU, ii
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Capítulo 2
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2.6: The s-waves travel slower, so
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2.14: (a) The displacement vector i
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2.18: a) The velocity at t = 0 is a
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2.20: a) The bumper’s velocity an
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2.26: a) x 0 < 0, v 0x < 0, a x < 0
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2.28: a) Interpolating from the gra
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2.31: a) At t = 3 s the graph is ho
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2.35: a) b) From the graph (Fig. (2
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2.37: a) The car and the motorcycle
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2.43: a) Using the method of Exampl
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2.45: a) v y = v0 y − gt = ( −6
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2.49: a) For a given initial upward
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2.51: a) From Eqs. (2.17) and (2.18
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2.53: a) The change in speed is the
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.0 m 2.56: a) 1.25m s. 25 = 20.0 s
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1 8 m ⎛ 365 d ⎞⎛ ⎞⎛ 2.62:
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2.66: a) The simplest way to do thi
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2 2.69: a) x = 1 2) a t , and with
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2.71: a) Travelling at 20 m s , Jua
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2.77: Let t 1 be the fall for the w
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2.82: a) 2.83: a) From Eq. (2.14),
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2 2.86: a) The helicopter accelerat
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2.90: a) Suppose that Superman fall
Page 89 and 90: 2 2.93: The velocities are vA = α
Page 91 and 92: 1 2.96: The time spent above y max
Page 93 and 94: 2.98: a) Let the height be h and de
Page 95 and 96: 3.1: a) v (5.3 m) − (1.1m) = 1.4
Page 97 and 98: 3.7: a) b) r ˆ ˆ 2 v = αi − 2
Page 99 and 100: 3.11: Take + y to be upward. Use Ch
Page 101 and 102: 3.15: a) Solving Eq. (3.17) for v =
Page 103 and 104: v y . g 2 9.80 m s 0 16.0 m s 3.18:
Page 105 and 106: 3.22: Substituting for t in terms o
Page 107 and 108: 3.25: Take + y to be downward. a) U
Page 109 and 110: 3.34: a) a 2 rad = ( 3 m/s) /(14 m)
Page 111 and 112: 3.45: a) The a = 0 and a y = −2β
Page 113 and 114: 3.48: a) The equations of motions a
Page 115 and 116: 3.52: a) Setting y = −h in Eq. (3
Page 117 and 118: 3.58: Equation 3.27 relates the ver
Page 119 and 120: 3.60: a) This may be done by a dire
Page 121 and 122: 3.64: Combining equations 3.25, 3.2
Page 123 and 124: 3.66: (a) v x ( runner) = vx (ball)
Page 125 and 126: 3.69: Take + y to be upward. a) The
Page 127 and 128: 3.73: a) The acceleration is given
Page 129 and 130: 3.76: a) dv dt = = d dt (1/ 2) vxax
Page 131 and 132: 3.85: The three relative velocities
Page 133 and 134: 3.90: As in the previous problem, t
Page 135 and 136: 3.92: The x-position of the plane i
Page 137 and 138: 4.1: a) For the magnitude of the su
Page 139: 2 4.11: a) During the first 2.00 s,
Page 143 and 144: 4.27: a) b) For the chair, a y = 0
Page 145 and 146: 4.33: a) The resultant must have no
Page 147 and 148: 2 4.39: a) Both crates moves togeth
Page 149 and 150: 4.43: a) The engine is pulling four
Page 151 and 152: 4.48: a) The net force on a point o
Page 153 and 154: 4.51: a) L is the lift force b) ∑
Page 155 and 156: 2 4.54: The velocity as a function
Page 157 and 158: 4.57: In this situation, the x-comp
Page 159 and 160: 5.1: a) The tension in the rope mus
Page 161 and 162: 5.14: a) In level flight, the thrus
Page 163 and 164: 2L 5.20: Similar to Exercise 5.16,
Page 165 and 166: 5.28: a) The stopping distance is 2
Page 167 and 168: 5.35: First, determine the accelera
Page 169 and 170: 5.40: Differentiating Eq. (5.10) wi
Page 171 and 172: 5.49: a) Setting arad = g in Eq. (5
Page 173 and 174: 5.56: The tension in the lower chai
Page 175 and 176: 5.63: (Denote F r by F.) a) The for
Page 177 and 178: 5.67: Consider the forces on the pe
Page 179 and 180: 5.70: It’s interesting to look at
Page 181 and 182: 5.72: The key idea in solving this
Page 183 and 184: 5.77: See Exercise 5.40. a) The max
Page 185 and 186: 5.82: We take the upward direction
Page 187 and 188: 5.85: Denote the magnitude of the a
Page 189 and 190: 5.92: (a) There is a contact force
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5.96: (a) 80 mi h is 35 .7 m s in S
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5.101: a) The rock is released from
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mg mg 5.103: Without buoyancy, kv t
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5.106: (a) To find find the maximum
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5.109: a) For the same rotation rat
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5.116: a) Differentiating twice, a
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5.120: a) There are many ways to do
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5.124: For convenience, take the po
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5.126: In all cases, the tension in
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6.1: a) ( 2.40 N) (1.5 m) = 3.60 J
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6.10: a) From Eq. (6.6), 1 2 5 ⎛
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6.18: As the example explains, the
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6.27: a) The friction force is µ k
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6.35: a) The static friction force
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6.42: The initial and final kinetic
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6.53: a) In terms of the accelerati
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6.58: The work per unit mass is ( W
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6.66: Let x be the distance past P.
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6.71: The velocity and acceleration
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6.79: In terms of the bumper compre
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6.85: f = .25 mg soW f = W = −(0.
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6.94: a) The number of cars is the
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(8.00 hp)(746 W/hp) 6.99: a) = 358
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6.101: a) Denote the position of a
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6.104: From F = m a, Fx = max, Fy =
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7.1: From Eq. (7.2), mgy = (800 kg)
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7.10: (a) The flea leaves the groun
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2 7.16: U = 1 ky , where y is the v
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1 2 1 2 7.24: From kx = mv , the re
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dU 3 3 7.34: From Eq. (7.15), F =
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7.39: a) At constant speed, the upw
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1 2 7.46: a) U − U = mg( h − 2R
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7.49: a) K 1 + U1 + Wother = K2 + U
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7.54: To be at equilibrium at the b
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7.57: The two design conditions are
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7.62: a) The skier’s kinetic ener
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1 2 1 2 7.69: With U = 0 , K = 0 K
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7.74: a) From either energy or forc
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2 2 2 2 7.78: a) a = d x dt = −ω
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3 7.83: a) Along this line, x = y ,
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7.86: a) The slope of the U vs. x c
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Capítulo 8
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8.7: ∆p ∆t ( 0.0450 kg)( 25.0 m
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8.14: The impluse imparted to the p
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8.21: a) Taking v A and v B to be m
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8.25: m + m )(3.00 m s) = m (4.00 m
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8.31: Use conservation of the horiz
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8.37: Let +y be north and +x be sou
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8.43: a) In Eq. (8.24), let mA = m
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8.52: It turns out to be more conve
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8.63: The total momentum must be ze
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8.69: (This problem involves solvin
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8.72: a) The stuntman’s speed bef
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8.76: Just after the collision: ∑
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8.80: a) From the derivation in Sec
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8.83: a) In terms of the primed coo
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8.86: (a) For momentum to be conser
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8.90: a) For the x - and y - direct
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8.96: The trick here is to realize
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8.98: The two fragments are 3.00 kg
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8.101: a) If the objects stick toge
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8.107: a) For t < 0 the rocket is a
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8.111: a) The tension in the rope a
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9.1: a) 1.50 m = 0.60 rad = 34.4°
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2 9.10: a) ω = ω + α t = 1.50 ra
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9.18: The following table gives the
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9.24: From a rad = ω 2 r, 4 which
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vr 5.00 m s 9.33: The angular veloc
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9.40: a) In the expression of Eq. (
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2π 2 2 9.49: a) ω = T , so Eq. (9
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9.60: For this case, dm = γ dx. a)
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dθ 2 2 3 2 9.65: a) ω z = = 2γt
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9.70: a) The angular acceleration w
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9.75: I approximate my body as a ve
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9.81: a) Consider a small strip of
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9.84: Taking the zero of gravitatio
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9.90: a) In the case that no energy
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9.94: a) From the parallel-axis the
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9.99: a) Following the hint, the mo
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Capítulo 10
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10.6: (a) τ A = (50 N)(sin 60° )(
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1 2 2 = m 2 8.25 kg 0.0750 m = 0.02
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10.23: n = mg cos α mg sin θ −
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10.26: a) The angular speed of the
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10.32: I = mL 1 2 1 2 2 = ( 117 kg)
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10.40: The skater’s initial momen
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10.46: (a) Conservation of angular
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2 2π rad 10.51: a) Solving Eq. (10
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2 2θ 2θ 2θ I 10.57: t = = = . α
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10.63: The net torque on the pulley
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2 10.67: For the disk, K = (3 4) Mv
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10.72: (a) Στ = Iα and a = Rα T
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10.75: a) Use conservation of energ
Page 377 and 378:
10.79: a) v = 10K 7m = ( 10)( 0.800
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10.84: (a) Στ = Iα 1 1 2 mg cos
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10.90: Angular momentum is conserve
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10.95: a) The initial angular momen
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10.100: a) The distance from the ce
Page 387 and 388:
10.102: Denoting the upward forces
Page 389 and 390:
11.1: Take the origin to be at the
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11.12: a) b) x = 6 .25 m when FA =
Page 393 and 394:
11.18: a) Denote the length of the
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11.30: a) The volume would increase
Page 397 and 398:
11.43: For the airplane to remain i
Page 399 and 400:
11.49: The horizontal component of
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11.56: (a) and (b) Lower rod: Στ
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11.59: a) ∑τ = 0, axis at lower
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−4 −4 2 11.64: a) ∆w = −σ
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11.68: (a) ΣτElbow = 0 F (3.80 cm
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11.71: a) Consider the forces on th
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11.74: a) The center of gravity of
Page 413 and 414:
11.79: a) Take torques about the po
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11.87: Use subscripts 1 to denote t
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11.93: a) Taking torques about the
Page 419 and 420:
11.95: Assume that the center of gr
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