Solution manual fundamentals of physics extended, 8th editionch07

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1 With speed v = 11200 m/s, we find K= mv = (2.9 × 105 ) (11200) = 18 × 1013 J 2 (a) The change in kinetic energy for the meteorite would be ( )( 1 ∆K = K f − K i = − K i = − mi vi2 = − × 106 kg 15 × 103 m/s 2 ) = −5 × 1014 J , or | ∆K |= × 1014 J The negative sign indicates that kinetic energy is lost (b) The energy loss in units of megatons of TNT would be § megaton TNT · −∆K = ( × 1014 J ) ă = 0.1megaton TNT 15 â 4.2 ì 10 J (c) The number of bombs N that the meteorite impact would correspond to is found by noting that megaton = 1000 kilotons and setting up the ratio: N= 0.1 × 1000 kiloton TNT = 13kiloton TNT (a) From Table 2-1, we have v = v02 + 2a∆x Thus, v = v02 + 2a∆x = ( 2.4 × 10 ) + ( 3.6 × 1015 ) ( 0.035) = 2.9 × 107 m/s (b) The initial kinetic energy is Ki = 2 mv0 = (1.67 × 10−27 kg )( 2.4 × 107 m/s ) = 4.8 × 10−13 J 2 The final kinetic energy is Kf = 2 mv = (1.67 × 10−27 kg )( 2.9 × 107 m/s ) = 6.9 × 10−13 J 2 The change in kinetic energy is ∆K = (6.9 × 10–13 – 4.8 × 10–13) J = 2.1 × 10–13 J We apply the equation x(t ) = x0 + v0t + 12 at , found in Table 2-1 Since at t = s, x0 = and v0 = 12 m/s , the equation becomes (in unit of meters) x(t ) = 12t + 12 at With x = 10 m when t = 1.0 s , the acceleration is found to be a = − 4.0 m/s The fact that a < implies that the bead is decelerating Thus, the position is described by x(t ) = 12t − 2.0t Differentiating x with respect to t then yields v(t ) = dx = 12 − 4.0t dt Indeed at t =3.0 s, v(t = 3.0) = and the bead stops momentarily The speed at t = 10 s is v(t = 10) = − 28 m/s , and the corresponding kinetic energy is K= mv = (1.8 × 10− kg)( − 28 m/s) = 7.1 J 2 We denote the mass of the father as m and his initial speed vi The initial kinetic energy of the father is Ki = Kson and his final kinetic energy (when his speed is vf = vi + 1.0 m/s) is K f = Kson We use these relations along with Eq 7-1 in our solution (a) We see from the above that Ki = 21 K f which (with SI units understood) leads to ê1 mvi = ô m ( vi + 1.0 ) » 2 ¬2 ¼ The mass cancels and we find a second-degree equation for vi : vi − vi − = 2 The positive root (from the quadratic formula) yields vi = 2.4 m/s b g F F mI I = G G Jv J K H2 H 2K (b) From the first relation above Ki = 21 Kson , we have mvi 2 son and (after canceling m and one factor of 1/2) are led to vson = 2vi = 4.8 m s By the work-kinetic energy theorem, W = ∆K = ( ) 2 mv f − mvi = (2.0 kg) (6.0 m/s) − (4.0 m/s) = 20 J 2 & & We note that the directions of v f and vi play no role in the calculation Eq 7-8 readily yields (with SI units understood) W = Fx ∆x + Fy ∆y = 2cos(100º)(3.0) + 2sin(100º)(4.0) = 6.8 J Using Eq 7-8 (and Eq 3-23), we find the work done by the water on the ice block: & & W = F ⋅ d = 210 ˆi − 150 ˆj ⋅ 15 ˆi − 12ˆj = (210) (15) + (−150) (−12) = 5.0 ×103 J ( )( ) Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as x = v0t + 21 at (where x0 = ) We choose to analyze the third and fifth points, obtaining 0.2 m = v0 (1.0 s) + a (1.0 s) 2 0.8m = v0 (2.0 s) + a (2.0 s) 2 Simultaneous solution of the equations leads to v0 = and a = 0.40 m s2 We now have two ways to finish the problem One is to compute force from F = ma and then obtain the work from Eq 7-7 The other is to find ∆K as a way of computing W (in accordance with Eq 7-10) In this latter approach, we find the velocity at t = 2.0 s from v = v0 + at (so v = 0.80 m s) Thus, W = ∆K = (3.0 kg) (0.80 m/s) = 0.96 J 10 The change in kinetic energy can be written as ∆K = 1 m(v 2f − vi2 ) = m(2a∆ x) = ma∆ x 2 where we have used v 2f = vi2 + 2a∆ x from Table 2-1 From Fig 7-27, we see that ∆ K = (0 − 30) J = − 30 J when ∆ x = + m The acceleration can then be obtained as a= ∆K (− 30 J) = = − 0.75 m/s m∆ x (8.0 kg)(5.0 m) The negative sign indicates that the mass is decelerating From the figure, we also see that when x = m the kinetic energy becomes zero, implying that the mass comes to rest momentarily Thus, v02 = v − 2a∆ x = − 2(− 0.75 m/s )(5.0 m) = 7.5 m /s , or v0 = 2.7 m/s The speed of the object when x = −3.0 m is v = v02 + 2a∆ x = 7.5 + 2(− 0.75)(− 3.0) = 12 = 3.5 m/s 66 The total weight is (100)(660) = 6.6 × 104 N, and the words “raises … at constant speed” imply zero acceleration, so the lift-force is equal to the total weight Thus P = Fv = (6.6 × 104)(150/60) = 1.65 × 105 W & 67 (a) The force F of the incline is a combination of normal and friction force which is serving&to “cancel” the tendency of the box to fall downward (due to its 19.6 N weight) & Thus, F = mg upward In this part of the problem, the angle φ between the belt and F is 80° From Eq 7-47, we have P = Fv cos φ = (19.6)(0.50) cos 80° = 1.7 W & (b) Now the angle between the belt and F is 90°, so that P = & (c) In this part, the angle between the belt and F is 100°, so that P = (19.6)(0.50) cos 100° = –1.7 W 68 Using Eq 7-7, we have W = Fd cos φ = 1504 J Then, by the work-kinetic energy theorem, we find the kinetic energy Kf = Ki + W = + 1504 J The answer is therefore 1.5 kJ 69 (a) In the work-kinetic energy theorem, we include both the work due to an applied force Wa and work done by gravity Wg in order to find the latter quantity ∆K = Wa + Wg Ÿ 30 = (100)(1.8) cos 180° + Wg leading to Wg = 2.1× 102 J (b) The value of Wg obtained in part (a) still applies since the weight and the path of the child remain the same, so ∆ Κ = Wg = 2.1× 102 J & 70 (a) To hold the crate at equilibrium in the final situation, F must have the same magnitude as the horizontal component of the rope’s tension T sin θ , where θ is the angle between the rope (in the final position) and vertical: θ = sin −1 FG 4.00IJ = 19.5° H 12.0 K But the vertical component of the tension supports against the weight: T cos θ = mg Thus, the tension is T = (230)(9.80)/cos 19.5° = 2391 N and F = (2391) sin 19.5° = 797 N An alternative approach based on drawing a vector triangle (of forces) in the final situation provides a quick solution (b) Since there is no change in kinetic energy, the net work on it is zero & & (c) The work done by gravity is Wg = Fg ⋅ d = − mgh , where h = L(1 – cos θ ) is the vertical component of the displacement With L = 12.0 m, we obtain Wg = –1547 J which should be rounded to three figures: –1.55 kJ (d) The tension vector is everywhere perpendicular to the direction of motion, so its work is zero (since cos 90° = 0) & (e) The implication of the previous three parts is that the work due to F is –Wg (so the net work turns out to be zero) Thus, WF = –Wg = 1.55 kJ & (f) Since F does not have constant magnitude, we cannot expect Eq 7-8 to apply 71 (a) Hooke’s law and the work done by a spring is discussed in the chapter Taking absolute values, and writing that law in terms of differences ∆F and ∆x , we analyze the first two pictures as follows: | ∆F | = k | ∆x | 240 N − 110 N = k (60 mm − 40 mm) which yields k = 6.5 N/mm Designating the relaxed position (as read by that scale) as xo we look again at the first picture: 110 N = k (40 mm − xo ) which (upon using the above result for k) yields xo = 23 mm (b) Using the results from part (a) to analyze that last picture, we find W = k (30 mm − xo ) = 45 N 72 (a) Using Eq 7-8 and SI units, we find & & W = F ⋅ d = (2 ˆi − ˆj) ⋅ (8 ˆi + c ˆj) = 16 − 4c which, if equal zero, implies c = 16/4 = m (b) If W > then 16 > 4c, which implies c < m (c) If W < then 16 < 4c, which implies c > m 73 A convenient approach is provided by Eq 7-48 P = F v = (1800 kg + 4500 kg)(9.8 m/s2)(3.80 m/s) = 235 kW Note that we have set the applied force equal to the weight in order to maintain constant velocity (zero acceleration) 74 (a) To estimate the area under the curve between x = m and x = m (which should yield the value for the work done), one can try “counting squares” (or half-squares or thirds of squares) between the curve and the axis Estimates between J and J are typical for this (crude) procedure (b) Eq 7-32 gives ´ a ¶1 x2 a a dx = – = J where a = –9 N·m2 is given in the problem statement 75 (a) Using Eq 7-32, the work becomes W = x2 – x3 (SI units understood) The plot is shown below: (b) We see from the graph that its peak value occurs at x = 3.00 m This can be verified by taking the derivative of W and setting equal to zero, or simply by noting that this is where the force vanishes (c) The maximum value is W = (3.00)2 – (3.00)3 = 13.50 J (d) We see from the graph (or from our analytic expression) that W = at x = 4.50 m (e) The case is at rest when v = Since W = ∆ K = mv / , the condition implies W = This happens at x = 4.50 m 76 The problem indicates that SI units are understood, so the result (of Eq 7-23) is in Joules Done numerically, using features available on many modern calculators, the result is roughly 0.47 J For the interested student it might be worthwhile to quote the “exact” answer (in terms of the “error function”): 1.2 ´ -2x² ¶.15 e dx = ¼ 2π [erf(6 /5) – erf(3 /20)] 77 (a) In 10 the cart moves mi ã Đ 5280 ft/mi ã Đ d = ă 6.0 ă (10 min) = 5280 ft h © 60 min/h ¹ © so that Eq 7-7 yields W = F d cos φ = (40 lb) (5280 ft) cos 30° = 1.8 × 105 ft ⋅ lb (b) The average power is given by Eq 7-42, and the conversion to horsepower (hp) can be found on the inside back cover We note that 10 is equivalent to 600 s Pavg 1.8 × 105 ft ⋅ lb = = 305 ft ⋅ lb / s 600 s which (upon dividing by 550) converts to Pavg = 0.55 hp 78 (a) Estimating the initial speed from the slope of the graph near the origin is somewhat difficult, and it may be simpler to determine it from the constant-acceleration equations from chapter 2: v = v0 + at and x = v0 + 21 at , where x0 = has been used Applying these to the last point on the graph (where the slope is apparently zero) or applying just the x equation to any two points on the graph, leads to a pair of simultaneous equations from which a = −2.0 m s and v0 = 10 m s can be found Then, K0 = mv0 = 2.5 × 103 J = 2.5 kJ (b) The speed at t = 3.0 s is obtained by v = v0 + at = 10 + (−2.0)(3.0) = 4.0 m/s or by estimating the slope from the graph (not recommended) Then the work-kinetic energy theorem yields W = ∆K = (50 kg)(4.0 m/s) − 2.5 ×103 J= − 2.1 kJ 79 (a) We set up the ratio FG H 50 km E = km megaton IJ K 1/ and find E = 503 ≈ × 105 megatons of TNT (b) We note that 15 kilotons is equivalent to 0.015 megatons Dividing the result from part (a) by 0.013 yields about ten million bombs 80 After converting the speed to meters-per-second, we find K = mv2 = 667 kJ ... that this agrees with the result of part (c) provides insight into the concept of work 12 (a) From Eq 7-6 , F = W/x = 3.00 N (this is the slope of the graph) (b) Eq 7-1 0 yields K = Ki + W = 3.00... [0.500 ∠ 30.0º]) = +1.31 J (b) Eq 7-1 0 (along with Eq 7-1 ) then leads to v = 2(1.31 J)/(3.00 kg) = 0.935 m/s 19 (a) We use F to denote the magnitude of the force of the cord on the block This force... factor of 1/2) are led to vson = 2vi = 4.8 m s 6 By the work-kinetic energy theorem, W = ∆K = ( ) 2 mv f − mvi = (2.0 kg) (6.0 m/s) − (4.0 m/s) = 20 J 2 & & We note that the directions of v f
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