The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want

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Dortha Miles
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1 8.38 (a) The mass moves down distance.0 m + x. Choose y = 0 at its lower point. K i + U gi + U si + E = K f + U gf + U sf 0 + mgy i = kx (.50 kg)9.80 m/s (.0 m + x) = (30 N/m) x 0 = (60 N/m)x (4.7 N)x 7.6 J x = 4.7 N ± ( 4.7 N) 4(60 N/m)( 7.6 N m) (60 N/m) 4.7 N ± 07 N x = 30 N/m The negative root tells how high the mass will rebound if it is instantly glued to the spring. We want x = 0.38 m

2 From the same equation, (.50 kg).63 m/s (.0 m + x) = (30 N/m) x 0 = 60x.44x.93 The positive root is x = 0.43 m (c) The full work-energy theorem has one more term: mgy i + fy i cos 80 = kx (.50 kg) 9.80 m/s (.0 m + x) N(.0 m + x) = (30 N/m) x 7.6 J N x J N x = 60 N/m x 60x 4.0x 6.8 = 0 x = 4.0 ± (4.0) 4(60)( 6.8) 30 x = 0.37 m 8.39 Choose U g = 0 at the level of the horizontal surface. Then E = (K f K i ) + (U gf U gi ) becomes: f s f x = (0 0) + (0 mgh) h or (µ k mg cos 30.0 ) sin 30.0ϒ mgh (µ k mg)x = h = 60.0 cm m = 3.00 kg θ = 30.0 Thus, the distance the block slides across the horizontal surface before stopping is: x = h µ k h cot 30.0 = h k cot 30.0ϒ = (0.600 m) cot 30.0ϒ 0.00 or x =.96 m

3 *8.40 The total mechanical energy of the diver is E mech = K + U g = mv + mgh. Since the diver has constant speed, de mech dt = mv dv dt + mg dh dt = 0 + mg( v) = mgv The rate he is losing mechanical energy is then de mech dt = mgv = (75.0 kg)(9.80 m/s )(60.0 m/s) = 44. kw 8.4 U(r) = A r F r = U r = d dr A r = A r 8.4 F x = ƒu ƒx = 3 ƒ(3x y 7x ƒx = (9x y 7) = 7 9x y F y = ƒu ƒy = 3 ƒ(3x y 7x ƒy = (3x 3 0) = 3x 3 Thus, the force acting at the point (x, y) is F = F x i + F y j = (7 9x y)i 3x 3 j *8.43 (a) There is an equilibrium point wherever the graph of potential energy is horizontal: At r =.5 mm and 3. mm, the equilibrium is stable. At r =.3 mm, the equilibrium is unstable. A particle moving out toward r approaches neutral equilibrium. The particle energy cannot be less than 5.6 J. The particle is bound if 5.6 J E < J. (c) If the particle energy is 3 J, its potential energy must be less than or equal to 3 J. Thus, its position is limited to 0.6 mm r 3.6 mm. (d) (e) K + U = E. Thus, K max = E U min = 3.0 J ( 5.6 J) =.6 J Kinetic energy is a maximum when the potential energy is a minimum, at r =.5 mm. (f) 3 J + W = J. Hence, the binding energy is W = 4 J.

4 *8.44 stable unstable neutral 8.45 (a) F x is zero at points A, C and E; F x is positive at point B and negative at point D. (c) A and E are unstable, and C is stable. F x B A C E x (m) D 8.46 (a) As the pipe is rotated, the CM rises, so this is stable equilibrium. As the pipe is rotated, the CM moves horizontally, so this is neutral equilibrium. (c) As the pipe is rotated, the CM falls, so this is unstable equilibrium. O CM O CM O CM a b c 8.47 (a) When the mass moves distance x, the length of each spring changes from L to x + L, so each exerts force k( x + L L) toward its fixed end. The y- components cancel out and the x-components add to: F x = k( x + L L) x = kx + x + L klx x + L

5 Choose U = 0 at x = 0. Then at any point U(x) = x Fx dx = x 0 0 kx + klx dx = k x + L x x xdx kl 0 0 x x + L dx U(x) = kx + kl(l x + L ) U(x) = 40.0x (.0 x +.44 ) U(x)(J) x(m) x, m U, J For negative x, U(x) has the same value as for positive x. The only equilibrium point (i.e., where F x = 0) is x = 0. (c) K i + U i + E = K f + U f J + 0 = mv f + 0 v f = J m 8.48 (a) E = mc = (9. χ 0 3 kg)( m/s) = J (c) (d) J J J 8.49 (a) est energy = mc = ( kg)( m/s) = J E = γmc = mc = J = J (v/c) (.990) (c) K = γmc mc = J J = J

6 8.50 The potential energy of the block is mgh. An amount of energy µ k mgs cos θ is lost to friction on the incline. Therefore the final height y max is found from where mgy max = mgh µ k mgs cos θ s = y max sin θ mgy max = mgh µ k mgy max cot θ Solving, h h y max = + µ k cot θ y max θ *8.5 m = mass of pumpkin = radius of silo top v i 0 n tθ mg v initially later F r = ma r n mg cos θ = m v When the pumpkin is ready to lose contact with the surface, n = 0. Thus, at the point where it leaves the surface: v = g cos θ. Choose U g = 0 in the θ = 90.0 plane. Then applying conservation of energy from the starting point to the point where the pumpkin leaves the surface gives K f + U gf = K i + U gi mv + mg cos θ = 0 + mg

7 Using the result from the force analysis, this becomes mg cos θ + mg cos θ = mg, which reduces to cos θ = 3, and gives θ = cos (/3) = 48.ϒ as the angle at which the pumpkin will lose contact with the surface. 8.5 (a) U A = mg = (0.00 kg)(9.80 m/s )(0.300 m) = J (c) v B = K A + U A = K B + U B K B = K A + U A U B = mg= J K B m = (0.588 J) 0.00 kg =.4 m/s A B C /3 (d) U C = mgh C = (0.00 kg)(9.80 m/s )(0.00 m) = 0.39 J K C = K A + U A U C = mg(h A h C ) K C = (0.00 kg)(9.80 m/s )( ) m = 0.96 J 8.53 (a) K B = mv B = (0.00 kg)(.50 m/s) = 0.5 J E = K + U = K B K A + U B U A = K B + mg(h B h A ) = 0.5 J + (0.00 kg)(9.80 m/s )( m) = 0.5 J J = J (c) It's possible to find an effective coefficient of friction, but not the actual value of µ since n and f vary with position. *8.54 v = 00 km/h = 7.8 m/s The retarding force due to air resistance is = DρAv = (0.330)(.0 kg/m3 )(.50 m )(7.8 m/s) = 38 N

8 Comparing the energy of the car at two points along the hill, K i + U gi + E = K f + U gf or K i + U gi + W e ( s) = K f + U gf where W e is the work input from the engine. Thus, W e = ( s) + (K f K i ) + (U gf U gi ) ecognizing that K f = K i and dividing by the travel time t gives the required power input from the engine as P = W e t = s t + mg y t = v + mgv sin θ P = (38 N)(7.8 m/s) + (500 kg)(9.80 m/s )(7.8 m/s)sin 3.0 P = 33.4 kw = 44.8 hp *8.55 At a pace I could keep up for a half-hour exercise period, I climb two stories up, forty steps each 8 cm high, in 0 s. My output work becomes my final gravitational energy, mgy = 85 kg(9.80 m/s )( m) = 6000 J making my sustainable power 6000 J 0 s = ~0 W 8.56 k = N/m m = 5.0 kg x A = 0.00 m U g x = 0 = U s x = 0 = 0 (a) E = K A + U ga + U sa = 0 + mgx A + kx A E = (5.0 kg)(9.80 m/s )( 0.00 m) + ( Ν/ µ )(0.00µ ) E = 4.5 J + 5 J = 00 J Since only conservative forces are involved, the total energy at point C is the same as that at point A. K C + U gc + U sc = K A + U ga + U sa 0 + (5.0 kg)(9.80 m/s )x C + 0 = J + 5 J x C = 0.40 J

9 (c) K B + U gb + U sb = K A + U ga + U sa (5.0 kg) v B = J + 5 J v B =.84 m/s Σ F = 0 (i.e., when the magnitude of the m upward spring force equals the magnitude of the downward gravitational force). This occurs at x < 0 where (d) K and v are at a maximum when a = kx = mg or x = (5.0kg)(9.80m / s Ν/ m ) = m Thus, K = K max at x = 9.80 mm (e) K max = K A + (U ga U g x = 9.80 mm ) + (U sa U s x = 9.80 mm ), or (5.0 kg) v max = (5.0 kg)(9.80 m/s )[( 0.00 m) ( m)] + ( N/m) [( 0.00 m) ( m) ] yielding v max =.85 m/s 8.57 E = W f E f E i = f d BC k x mgh = µmgd mgh k x µ = mgd = 0.38 A 3.00 m 6.00 m B C

10 Goal Solution G: We should expect the coefficient of friction to be somewhere between 0 and since this is the range of typical µ k values. It is possible that µ k could be greater than, but it can never be less than 0. O: The easiest way to solve this problem is by considering the energy conversion for each part of the motion: gravitational potential to kinetic energy from A to B, loss of kinetic energy due to friction from B to C, and kinetic to elastic potential energy as the block compresses the spring. Choose the gravitational energy to be zero along the flat portion of the track. A: Putting the energy equation into symbols: U ga W BC = U sf expanding into specific variables: mgy A f d BC = kx s where f = µ mg solving for the unknown variable: µ mgd = mgy kx or µ = y d kx mgd substituting: µ = 3.00 m 6.00 m (50 N/m)(0.300 m) (0.0 kg)(9.80 m/s )(6.00 m) = 0.38 L: Our calculated value seems reasonable based on the friction data in Table 5.. The most important aspect to solving these energy problems is considering how the energy is transferred from the initial to final energy states and remembering to subtract the energy resulting from any non-conservative forces (like friction) The nonconservative work (due to friction) must equal the change in the kinetic energy plus the change in the potential energy. Therefore, µ k mgx cos θ = K + kx mgx sin θ and since v i = v f = 0, K = 0. Thus, µ k (.00)(9.80)(cos 37.0 )(0.00) = (00)(0.00) (.00)(9.80)(sin 37.0 )(0.00) and we find µ k = 0.5. Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy. The net loss in mechanical energy is equal to the energy lost due to friction.

11 8.59 (a) Since no nonconservative work is done, E = 0 Also K = 0 therefore, U i = U f where U i = (mg sin θ)x and U f = kx Substituting values yields (.00)(9.80) sin 37.0 = (00) x and solving we find x = 0.36 m F = ma. Only gravity and the spring force act on the block, so kx + mg sin θ = ma For x = 0.36 m, a = 5.90 m/s The negative sign indicates a is up the incline. The acceleration depends on position. (c) U(gravity) decreases monotonically as the height decreases. U(spring) increases monotonically as the spring is stretched. K initially increases, but then goes back to zero. *8.60 (a) F = d dx ( x3 + x + 3x)i = (3x 4x 3)i F = 0 when x =.87 and (c) The stable point is at x = point of minimum U(x) 5 The unstable point is at x =.87 maximum in U(x) F(x) U(x) 5

12 8.6 (K + U) i = (K + U) f 0 + (30.0 kg)(9.80 m/s )(0.00 m) + (50 N/m)(0.00 m) = (50.0 kg) v + (0.0 kg)(9.80 m/s )(0.00 m) sin J J = (5.0 kg)v + 5. J v =.4 m/s 30.0 kg 0.0 kg 0.0 cm (a) Between the second and the third picture, E = K + U m k µmgd = mv i + kd v i (50.0 N/m) d (.00 kg)(9.80 m/s )d (.00 kg)(3.00 m/s ) = 0 d v f = 0 [.45 ±.35] N d = 50.0 N/m = m Between picture two and picture four, E = K + U v f(d) = mv + mv i v = (3.00 m/s) (.45 N)()(0.378 m) (.00 kg) =.30 m/s v = 0 D

13 (c) For the motion from picture two to picture five, E = K + U f(d + d) = (.00 kg)(3.00 m/s) 9.00 J D = (0.50)(.00 kg)(9.80 m/s ) (0.378 m) =.08 m 8.63 (a) T v T v B B m x k Initial compression of spring: kx = mv (450 N/m)( x) = (0.500 kg)(.0 m/s) x = m Speed of block at top of track: E = W f mgh T + mv T mgh B + mv B = f(π) (0.500 kg)(9.80 m/s )(.00 m) + (0.500 kg) v T (0.500 kg)(.0 m/s) = (7.00 N)(π)(.00 m) 0.50v T = 4. v T = 4.0 m/s (c) Does block fall off at or before top of track? Block falls if a r < g a r = v T = (4.0).00 = 6.8 m/s therefore a r > g and the block stays on the track.

14 8.64 Let λ represent the mass of each one meter of the chain and T represent the tension in the chain at the table edge. We imagine the edge to act like a frictionless pulley. (a) For the five meters on the table with motion impending, F y = 0 + n 5λg = 0 n f T T 5λg T T P n = 5λg f s µ s n = 0.6(5λg) = 3λg 3λg F x = 0 + T f s = 0 T = f s T The maximum value is barely enough to support the hanging segment according to F y = 0 + T 3λg = 0 T = 3λg so it is at this point that the chain starts to slide. Let x represent the variable distance the chain has slipped since the start. Then length (5 x) remains on the table, with now F y = 0 + n (5 x)λg = 0 n = (5 x)λg f k = µ k n = 0.4(5 x)λg = λg 0.4xλg Consider energies at the initial moment when the chain starts to slip, and a final moment when x = 5, when the last link goes over the brink. Measure heights above the final position of the leading end of the chain. At the moment the final link slips off, the center of the chain is at y f = 4 meters. Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at height 8 3 = 6.5 m. 3λg K i + U i + E = K f + U f 0 + (m gy + m gy ) i + i f f k dx cos θ = mv + mgy f (5λg)8 + (3λg) (λg 0.4xλg) dx cos 80 = (8λ) v + (8λg)4

15 40.0 g g.00 g dx g 0 x dx = 4.00v g 7.5 g.00 gx g x 5 0 = 4.00v 7.5 g.00 g(5.00) g(.5) = 4.00v.5 g = 4.00v v = (.5 m)(9.80 m/s ) 4.00 = 7.4 m/s 8.65 (a) On the upward swing of the mass: K i + U i + E = K f + U f v i θ) mv i = 0 + mgl( cos L m θ v i = gl( cos θ) v i = (9.80 m/s )(.0 m)( cos 35.0 ) (a) v i =.06 m/s 8.66 Launch speed is found from mg v = 4 5 h = mv g 4 5 h h θ h/5 y v y = v sin θ The height y above the water (by conservation of energy) is found from mgy = mv y + mg h 5 since mv x is constant in projectile motion y = g v y + h 5 = g v sin θ + h 5 y = g g 4 5 h sin θ + h 5 = 4 5 h sin θ + h 5

16 8.67 (a) Take the original point where the ball is released and the final point where its upward swing stops at height H and horizontal displacement F Pivot L F Pivot L x = L (L H) = LH H Since the wind force is purely horizontal, it does work (a) m m H W wind = F ds = F dx = F LH H [The wind force potential energy change would be F LH H ] The work-energy theorem can be written: K i + U gi + W wind = K f + U gf, or F LH H = 0 + mg H giving F LH F H = m g H Here H = 0 represents the lower turning point of the ball's oscillation, and the upper limit is at F (L) = (F + m g )H. Solving for H yields LF H = F + m g = L + (mg/f) As F 0, H 0 as is reasonable. As F, H L, which is unreasonable. (.00 m) H = + [(.00 kg)(9.80 m/s )/4.7 N] =.44 m (c) Call θ the equilibrium angle with the vertical. ΣF x = 0 T sin θ = F, and ΣϖF y = 0 T cos θ = mg Dividing: tan θ = F mg = 4.7 N 9.6 N = 0.750, or θ = 36.9 Therefore, H eq = L( cos θ) = (.00 m)( cos 36.9 ) = m (d) As F, tan θ, θ 90.0 and H eq ϖϖ L A very strong wind pulls the string out horizontal, parallel to the ground. Thus, (H eq ) max = L

17 8.68 Call φ = 80 θ the angle between the upward vertical and the radius to the release point. Call v r the speed here. By conservation of energy K i + U i + E = K r + U r mv i + mg + 0 = mv r + mg cos φ g + g = v r + g cos φ v r = 3 g g cos φ The path after string is cut m θ C v i = g The components of velocity at release are v x = v r cos φ and v y = v r sin φ so for the projectile motion we have x = v x t sin φ = v r cos φ t y = v y t gt cos φ = v r sin φ t gt By substitution cos φ = v r sin φ sin φ v r cos φ g with sin φ + cos φ =, sin φ v r cos φ g sin φ = v r cos φ = cos φ(3 g g cos φ) sin φ = 6 cos φ 4 cos φ = cos φ 3 cos φ 6 cos φ + = 0 6 ± 36 cos φ = 6 Only the sign gives a value for cos φ that is less than one: cos φ = φ = so θ = 00.6

18 8.69 Applying Newton's second law at the bottom and top (t) of the circle gives v t T b mg = mv b and T t mg = mv t mg T t Adding these gives T b = T t + mg + m(v b v t) T b Also, energy must be conserved and U + K = 0 So, m(v b v t) + (0 mg) = 0 and m(v b v t) = 4mg mg v b Substituting into the above equation gives T b = T t + 6mg 8.70 (a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball's speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides. elative to the point of suspension, L θ d U i = 0, U f = mg[d (L d)] Peg From this we find that mg(d L) + mv = 0 Also for centripetal motion, mg = mv where = L d. Upon solving, we get d = 3L 5

19 8.7 (a) The potential energy associated with the wind force is +Fx, where x is the horizontal distance traveled, with x positive when swinging into the wind and negative when swinging in the direction the wind is blowing. The initial energy of Jane is, (using the pivot point of the swing as the point of zero gravitational energy), E i = (K + U g + U wind ) i = mv i mgl cos θ FL sin θ where m is her mass. At the end of her swing, her energy is E f = (K + U g + U wind ) f = 0 mgl cos φ + FL sin φ so conservation of energy (E i = E f ) gives mv i mgl cos θ FL sin θ = mgl cos φ + FL sin φ This leads to v i = gl(cos θ cos φ) + FL m (sin θ + sin φ) But D = L sin φ + L sin θ, so that sin φ = D L sin θ = sin 50.0 = which gives φ = 8.9. Using this, we have v i = 6.5 m/s. Here (again using conservation of energy) we have, MgL cos φ + FL sin φ + Mv = MgL cos θ FL sin θ where M is the combined mass of Jane and Tarzan. Therefore, v = gl(cos φ cos θ) FL (sin φ + sin θ) which gives v = 9.87 m/s M as the minimum speed needed. 8.7 Find the velocity at the point where the child leaves the slide, height h: (U + K) i = (U + K) f mgh + 0 = mgh + mv v = g(h h) H Use Newton's laws to compare h and H. θ

20 (ecall the normal force will be zero): F r = ma r = mv mg sin θ n = mv m( g)(h h) mg sin θ = Put θ in terms of : sin θ = h mg h = mg(h h) h = 3 H Notice if H 3, the assumption that the child will leave the slide at a height H is no 3 longer valid. Then the velocity will be too large for the centripetal force to keep the child on the slide. Thus if H Case I: Surface is frictionless mv = kx, the child will leave the track at h =. k = mv (5.00 kg)(.0 m/s) x = 0 m = N/m Case II: Surface is rough, µ k = mv = kx µ k mgx 5.00 kg v = (7.0 0 Ν/ m)(0 m) (0.300)(5.00 kg)(9.80 m/s )(0 m) v = 0.93 m/s 8.74 ΣF y = n mg cos 37.0 = 0, n = mg cos 37.0 = 400 N f = µn = (0.50)(400) = 00 N W f = E ( 00)(0.0) = U A + U B + K A + K B U A = m A g(h f h i ) = (50.0)(9.80)(0.0 sin 37.0 ) =

21 U B = m B g(h f h i ) = (00)(9.80)( 0.0) = K A = m A(v f v i ) K B m B(v f v i ) = m B m A K A = K A Adding and solving, K A = 3.9 kj

P8.14. m 1 > m 2. m 1 gh = 1 ( 2 m 1 + m 2 )v 2 + m 2 gh. 2( m 1. v = m 1 + m 2. 2 m 2v 2 Δh determined from. m 2 g Δh = 1 2 m 2v 2.

P8.14. m 1 > m 2. m 1 gh = 1 ( 2 m 1 + m 2 )v 2 + m 2 gh. 2( m 1. v = m 1 + m 2. 2 m 2v 2 Δh determined from. m 2 g Δh = 1 2 m 2v 2. . Two objects are connected by a light string passing over a light frictionless pulley as in Figure P8.3. The object of mass m is released from rest at height h. Using the principle of conservation of