Physics 40 Chapter 7 Homework Solutions

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1 Phsics 40 Chapter 7 Homework Solutions T = F 3 g (1) T sin θ + T sin θ = Fg () 1 1 T cosθ = T cosθ (3) 1 1 Eliminate T and solve for T 1 Fgcos θ T = = T 1 3 g ( sin θ cosθ + cosθ sin θ ) sin ( θ + θ ) = F = g N cos5.0 T1 = Fg = sin N F cosθ cosθ 1 cos60.0 T = T1 = 96 N = 163 N cosθ cos5.0 See ke in glass case.

2 F= ma reads (.00iˆ+.00ˆj+ 5.00iˆ 3.00ˆj 45.0 iˆ) N = m( 3.75 m s ) a ˆ iˆ ˆj a ˆ where â represents the direction of a ( ) N = m( 3.75 m s ) F = ( 4.0 ) + ( 1.00 ) N at tan 4.0 F= 4.0 N at 181 =m 3.75 m s a. ˆ below the -ais For the vectors to be equal, their magnitudes and their directions must be equal. (a) a ˆ is at 181 counterclockwise from the -ais 4.0 N (b) m = = 3.75 m s 11. kg (d) f = i + t = + ( ) v v a m s at s so v f = 37.5 m s at 181 v 37.5m s cos181 ˆ 37.5m s sin 181 ˆ f = i + j so v ˆ ˆ f = ( 37.5i j ) m s (c) v f = m s = 37.5 m s (Case 1, impending upward motion) Setting F P n fs, ma µ sn fs, ma µ sp Setting = 0 : cos50.0 = 0 = : = cos50.0 ma = P= 0.161P F = 0 : Psin P = 0 P = 48.6 N (Case, impending downward motion) As in Case 1, f, ma = 0.161P Setting min s F = 0 : Psin P = 0 P = 31.7 N

3 Choose the -ais pointing down the slope. a = 5.00 m s Consider forces on the to. v = v + at : 30.0 m s = 0 + a 6.00 s f i a = 5.00 m s. F = ma : mg sin θ = m 5.00 m s θ = 30.7 F = ma : mg cosθ + T = 0 T = mgcosθ = cos30.7 T = N See Ke in Glass Case

4 Knight Book Problems From Chapter 6: 8, 15, 6, 7, 33, 43, Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and s, v 1 m/s 0 s a = = = 6 m/s t s For t between 3 s and 6 s, v = 0 m/s, so a = 0 m/s. For t between 6 s and 8 s, v 0 m/s 1 m/s a = = = 4 m/s t 3 s From Newton s second law, at t = 1 s we have Fnet = ma = (.0 kg)(6 m/s ) = 1 N At t = 4 s, a = 0 m/s, so F net = 0 N. At t = 7 s, Fnet = ma = (.0 kg)( 4.0 m/s ) = 8 N Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down Model: We assume that the passenger is a particle acted on b onl two vertical forces: the downward pull of gravit and the upward force of the elevator floor. Visualize: The graph has three segments corresponding to different conditions: (1) increasing velocit, meaning an upward acceleration; () a period of constant upward velocit; and (3) decreasing velocit, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then appl Equation The acceleration for the first segment is v1 v0 8 m/s 0 m/s a = = = 4 m/s t t s 0 s 1 0

5 a 4 m/s 4 w = mg 1+ = mg 1 + = (75 kg)(9.80 m/s ) 1+ = 1035 N g 9 80 m/s For the second segment, a = 0 m/s and the weight is 0 m/s w = mg 1 + = mg = (75 kg)(9.80 m/s ) = 740 N g For the third segment, v3 v 0 m/s 8 m/s a = = = m/s t3 t 10 s 6 s m/s w = mg 1 + = (75 kg)(9.80 m/s )(1 0.) = 590 N 9.80 m/s Assess: As epected, the weight is greater than the gravitational force on the passenger when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is the gravitational force. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator. We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was calculated as follows: Fma 10 N ama = = = m/s m 5 kg Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more general result that vt () = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v 0 = 0 m/s. The area under the acceleration curve between 0s and 6s is 1 (4 s)( m/s ) = 4.0 m/s. We ve used the fact that the area between 4s and 6s is zero. Thus, at t = 6 s, v = 4. 0 m/s Visualize: The acceleration is a = F / m, so the acceleration-versus-time graph has eactl the same shape as the force-versus-time graph. The maimum acceleration is a ma = F ma / m= (6 N)/( kg) = 3 m/s. Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more general result that vt () = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v 0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle (3 m/s s 6 m/s) 1 3 m/s s = 3 m/s. Thus v = 9 m/s at t = 4 s. = plus a triangle

6 6.33. Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravit, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulles are frictionless, the tension is the same everwhere in the rope. Because all pulles are in equilibrium, their net force is zero. So the do not contribute to T. Visualize: Solve: (a) From the free-bod diagram for the mass, the tension in the rope is T = FG = mg = (6 kg)(9.80 m/s ) = 58.8 N 59 N (b) Using Newton s first law for the vertical direction on the pulle attached to the foot, ( F ) = F = Tsinθ Tsin15 ( F ) = 0 N net G foot Tsin15 + ( FG ) foot mfootg (4 kg)(9.80 m/s ) sinθ = = sin15 + = = = 0.96 T T 58.8 N θ = sin 0.96 = (c) Using Newton s first law for the horizontal direction, ( F ) = F = Tcosθ + Tcos15 F = 0 N net 1 traction Ftraction = Tcosθ + Tcos15 = T(cos cos15 ) = (58.8 N)( ) = (58.8 N)(1.344) = 79 N Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of the traction force, roughl one-tenth of the gravitational force on a human bod, seems reasonable.

7 6.43. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravit, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize: Solve: From the height of the slope and its angle, we can calculate its length: h h 50 m = sinθ 1 0 = = = 88 m sinθ sin Since Sam is not accelerating in the -direction, we can use Newton s first law to calculate the normal force: ( F ) = F = n F cosθ = 0 N n = F cosθ = mg cos θ = (75 kg)(9.80 m/s )(cos10 ) = 74 N net G G One-dimensional kinematics gives us Sam s acceleration: v1 v0 (40 m/s) 0 m /s v1 = v0 + a( 0) a = = =.78 m/s ( ) (88 m) Then, from Newton s second law and the equation f k = µ k n: 1 ( F ) = F = F sinθ + F f = ma net G thrust k mg sin θ + Fthrust ma (75 kg)(9.80 m/s )(sin10 ) + 00 N (75 kg)(.78 m/s ) µ k = = = n 74 N Assess: This coefficient seems a bit high for skis on snow, but not impossible.

8 6.64. Model: Model the object as a particle. The acceleration is not constant so we can t use the kinematic equations. All the motion is in the -direction. Visualize: Divide F b m to get a and then integrate twice. The constants of integration are both zero because of the initial conditions. F F0 tt / Solve: a() t = = ( e ) m m (a) t t F 0 F 0 v () T T t = adt e dt T e = = + C m m F0 The constant of integration is not zero. v(0) = 0 C = ( T) m t t F 0 F0 F 0 v T 1 T t = T e + T = T e m m m (b) After a ver long time the decaing eponential term is close to zero so 0 () F v t T m. Assess: It seems reasonable that the velocit after time T would increase with T and that the position at time T would increase with T Model: Use the linear model of drag. Assume the microorganisms are swimming in water at 0 C. 3 Visualize: The viscosit of water is η = N s/m at 0 C. Solve: (a) F = Fprop D = 0 Fprop = 6πηRv For a paramecium Fprop = 6 π ( N s/m )(50 10 m)( m/s) = N For an E. coli bacterium Fprop = 6 π ( N s/m )( m)(30 10 m/s) = N (b) Fprop Fprop Fprop a = = = m ρv 4 ρ πr 3 For a paramecium N a = = 1.8 m/s (1000 kg/m ) π (50 10 m) 3 For an E. coli bacterium N a = = 135 m/s (1000 kg/m ) π ( m) 3 Assess: The two accelerations are within a factor of two of each other.

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