Chapter 5 Newton Law Practice Proble Solution Student Textbook page 163 1. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the object. - The net force i the only horizontal force on the object. - The net force deterine the acceleration of the object according to Newton econd law of otion. F N F net = 2.2 N[E] F g The acceleration of the object Variable and Contant Known Iplied Unknown = 4.0 kg a v F net = 2.2 N[E] Strategy Since the net force i known, ue Newton econd law in ter of acceleration. Ue 1 N kg = 1. 2 The acceleration of the object i 0.55 / 2 [E]. Calculation v v Fnet a = v 22. N[E] a = 40. kg v a = 055. N[E] 2 Validate The object accelerated in the direction of the net force. The unit for acceleration i / 2 which i correct. 80
2. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the object. - The friction force i in the oppoite direction of the applied force. - The net force i the u of the friction force and the applied force on the object. - The net force deterine the acceleration of the object according to Newton econd law of otion. The acceleration of the object Variable and Contant Known Iplied Unknown = 6.0 kg a F a = 4.4 N[E] F net F f = 1.2 N[E] Strategy Since the otion i all along one line, eat and wet, denote direction with ign only. Let eat be poitive and wet be negative. Find the net force on the object by finding the vector u of the applied force and the friction force acting on the object. Apply Newton econd law in ter of acceleration and olve. Ue 1 N kg = 1. 2 The acceleration of the object i 0.53/ 2 [E]. Calculation F net = F a + F f F net =+4.4 N 1.2 N F net =+3.2 N a = F net a = +3.2 N 6.0 kg a =+0.53 2 a = 0.53 [E] 2 Validate The object accelerated in the direction of the net force. The unit for acceleration i / 2 which i correct. 3. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the wheeled cart. - Since frictional force are ignored the net force on the wheeled cart i fro the tretched elatic band only. - The a of the cart i related to the net force on the cart and to the acceleration of the cart. 81
F N F net = 2.5 N[E] F g The a of the cart Variable and Contant Known Iplied Unknown F a = 2.5 N[E] F net a = 1.5 [E] 2 Strategy Calculation Find the net force on the wheeled cart. F net = F a F net = 2.5 N[E] Apply Newton econd law in ter of a and olve. Since 1 N kg = 1 it follow that 2 1N= 1kg and the unit for a i kg. 2 The a of the cart i 1.7 kg. = F net a = 2.5 N[E] 1.5 2 [E] = 1.67 kg = 1.7 kg Validate The unit for a of the wheeled cart i kg, which i correct. Practice Proble Solution Student Textbook page 168 4. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the object. - The friction force i in the oppoite direction of the applied force. - The net force i the u of the friction force and the applied force on the object. - The net force deterine the acceleration of the object according to Newton econd law of otion. - The equation of otion for unifor acceleration apply to the otion of the tone. 82
F N F f = 2.5 N[S] F a = 5.5 N[N] F g How far the object travelled Variable and Contant Known Iplied Unknown = 15 kg vi = 0.0 a F a = 5.5 N[N] F net F f = 2.5 N[S] t = 4.0 Strategy Since the otion i all along one line, north and outh, denote direction with ign only. Let north be poitive and outh be negative. Find the net force on the object by finding the vector u of the applied force and the friction force acting on the object. Apply Newton econd law in ter of acceleration and olve. 1 N kg = 1. 2 d Calculation F net = F a + F f F net =+5.5 N 2.5 N F net =+3N Recall the equation of otion fro a = F net a = +3N 15 kg Chapter 2. Ue the equation that a =+0.2 relate initial velocity, tie, acceleration 2 and diplaceent. a = 0.2 [N] 2 Subtitute in the known value and d = v i t + 1 2 a t 2 olve for d. d = 0.0 + 1 2 (0.2 )(4.0 ) 2 The object travelled 1.6 [N] after 4.0. 2 d = (0.1)(16.0 ) d = 1.6 d = 1.6 [N] Validate The object accelerated in the direction of the net force, which wa north, and travelled in that direction. The unit for diplaceent i, which i correct. 83
5. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the cyclit. - The friction force i in the oppoite direction of the cyclit applied force. - The net force i the u of the friction force and the applied force on the cyclit. - The net force deterine the acceleration of the cyclit according to Newton econd law of otion. - The equation of otion for unifor acceleration apply to the otion of the cyclit. F N F f = 50.0 N[W] F a = 325 N[E] F g (a) The acceleration of the cyclit (b) How far the tudent travelled Variable and Contant Known Iplied Unknown F a = 325 N[E] vi = 3.0 [E] a F f = 50 N[W] t = 8.0 T = 49 kg F net d Strategy Find the total a of the cyclit and the bicycle. Since the otion i all along one line, eat and wet, denote direction with ign only. Let eat be poitive and wet be negative. Find the net force on the cyclit by finding the vector u of the applied force and the friction force acting on the cyclit. Apply Newton econd law in ter of acceleration and olve. 1 N kg = 1. 2 Recall the equation of otion fro Chapter 2. Ue the equation that relate initial velocity, tie, acceleration and diplaceent. Calculation (a) T = 45 kg + 4.0 kg T = 49.0 kg F net = F a + F f F net =+325 N 50.0 N F net =+275 N a = F net a = +275 N 49.0 kg a =+5.6 2 a =+5.6 2 [N] 84
Subtitute in the known value and (b) d = v i t + 1 2 a t 2 olve for d. d = 3.0 (8.0 ) + 1 2 (+5.6 )(8.0 ) 2 2 d =+24 + (2.8)(64.0 ) d =+24 + 179.2 d =+203.2 d = 2.0 203 [E] 10 2 [E] (a) The acceleration of the cyclit i 5.6 / 2 [E]. (b) The cyclit travelled 2.0 10 2 [E] after 8.0. Validate The object accelerated in the direction of the net force, which wa north, and travelled in that direction. The unit for acceleration and diplaceent hould be / 2 and, repectively, and they are, which i correct. 6. Frae the Proble - Draw a free body diagra of the proble. - The downward force of gravity i balanced by the upward noral force. Therefore, there i no net force in the vertical direction. Thee force do not affect the acceleration of the car. - The friction force i the only force acting on the car that will low it down, o thi i the net force. - The a of the car i related to the net force on the car and to the acceleration of the car. - Ue the equation of otion fro Chapter 2 to olve for the acceleration. F N F f N E F g The coefficient of friction between the road and the car tire. Variable and Contant Known Iplied Unknown = 1.2 10 3 kg g = 9.8 F 2 net F f vi = 45 k/h a F N d = 35 F g vf = 0 k/h 85
Strategy Calculation Convert k/h to /. 45 k h 1h 3600 To find the friction force acting on the car the acceleration ut be known o that a relationhip between the net force and the friction force can be obtained. a = 156.25 ( )2 2(35 ) a = 156.25 ( )2 70 1000 1k Ue the equation of otion that relate a = v2 f vi 2 2d initial and final velocity acceleration and a = (0 )2 (12.5 diplaceent. )2 2d Find the net force on the car, knowing it a and acceleration uing Newton econd law. = 12.5 Ue the equation for weight to find the weight and thu the noral force. Since 1 N kg = 1 it follow that 2 1N= 1kg and the unit for a i kg. 2 Apply the equation for a frictional force. a = 2.23 2 F net = a = (1.2 10 3 kg)(2.23 ) 2 F net = 2678 N F f = µ k F N F N = g = (1.2 10 3 kg)(2.23 N kg ) = 11772 N F net = F f F net = µ k F N 2678.6 N = µ k 11772 N µ k = 2678.6 N 11772 N = 0.23 The coefficient of friction between the lippery road and the car tire i 0.23. Validate The coefficient of kinetic friction i low and it hould be ince the urface i a lippery road. 7. Conceptualize the Proble - The ion are oving horizontally (the cae in all linear accelerator), under an electric force. - Due to the ion all a, the force of gravity on the ion i exceedingly all. And, a indicated, the tie of flight i very hort. Therefore, the effect of the force of gravity i too all to be detected and you can conider the electric force to be the only force affecting the ion. How the force work doen t atter; all that atter i that it accelerate the ion in a horizontal direction. - No frictional force are preent. - The ion i initially at ret, eaning that the initial velocity i zero. - Before finding the force, the acceleration can be found uing the kineatic data. - Newton econd law can be applied to find the force fro the a and acceleration. - Let the direction of the force, and therefore the direction of the acceleration, be poitive. The agnitude of the force, F, required to accelerate the ion 86
Identify the Variable Known Iplied Unknown ion = 7.2 10 25 kg v 1 = 0 / F v 2 = 7.3 10 6 / a t = 5.5 10 6 Develop a Strategy Apply an appropriate kineatic equation to find the agnitude of the acceleration. Calculation v 2 = v 1 + at Subtitute and olve. a = v 2 v 1 t a = 7.3 106 / 0 / 5.5 10 6 a = 1.327 10 12 / 2 Apply Newton econd law to find the agnitude of the force. The agnitude of the force i 9.6 10 13 N in the poitive direction. a 1.3 10 12 / 2 F = a F = (7.2 10 25 kg)(1.327 10 12 / 2 ) F = 9.56 10 13 N F 9.6 10 13 N Validate the Solution Though the acceleration of the ion i relatively high, 1.3 10 12 / 2, the a of the ion i very all, o we expect that even a all force can produce thi acceleration. 8. Conceptualize the Proble - While the hockey tick i in contact with the hockey puck, it exert an average force and accelerate the puck. - The puck accelerate in the direction of the applied force. Define thi direction a poitive. - The puck accelerate fro ret (v 1 = 0 /) to oe unknown final velocity (v 2 =?) over the ditance, d, that it in contact with the hockey tick. - It i aued that friction between the puck and ice i negligible, o all of the applied force goe into accelerating the puck. The final velocity, v 2, of the hockey puck Identify the Variable Known Iplied Unknown F = 39 N v 1 = 0 / a = 0.20 kg v 2 d = 0.22 = 0.20 kg 87
Develop a Strategy Apply Newton econd law to find the agnitude of the net force. Write Newton econd law in ter of acceleration. Subtitute and olve. N kg i equivalent to 2 Apply an appropriate kineatic equation that relate initial velocity, acceleration and diplaceent to final velocity. Calculation F = a a = F a = 39 N 0.22 kg a = 195 2 v2 2 = v2 1 + 2ad v2 2 = 0 + 2(195 )(0.22 ) 2 v 2 = 9.2628 v 2 93. The final velocity of the puck i about 9.3 / in the direction of the applied force. Validate the Solution The acceleration i poitive, a expected, becaue it i in the ae direction a the applied force. The final velocity i 9.3 /, which i about 33 k/h, which i reaonable for a pa to a tea-ate, but uch too low for a laphot. Practice Proble Solution Student textbook page 170 171 9. Conceptualize the Proble - Begin by drawing a free-body diagra. - Motion of the led i in the horizontal direction, o the net horizontal force i cauing the led to accelerate. - Let the direction of the otion be the poitive horizontal direction. - There i no otion in the vertical direction, o the vertical acceleration i zero. If the acceleration i zero, the net vertical force ut be zero. The inforation lead to the value of the noral force. Let up be the poitive vertical direction. - Once the noral force i known, the force of friction can be deterined. - The led i not initially at ret, o the coefficient of kinetic friction applie to the otion. The acceleration, a, of the led Identify the Variable Known Iplied Unknown F = 1.3 10 3 N a y = 0 / 2 a x = 1.1 10 4 kg µ = 0.80 F N Develop a Strategy To find the noral force, apply Newton econd law to the vertical force. Analyze the free-body Calculation F N g = a y F N g = 0 F N = g 88
diagra to find all of the vertical force that act on the crate. To find the acceleration, apply Newton econd law to the horizontal force. Analyze the free-body diagra to find all of the horizontal force that act on the crate. Subtitute and olve. F F f = a x F f = µf N = µg F µg = a x a x = F µg a x = 1.3 103 N (0.8)(1.1 10 4 kg)(9.81 / 2 ) 1.1 10 4 kg N a x = 7.73 / 2 kg i equivalent to. 2 a x 7.7 / 2 The acceleration of the led i 7.7 / 2 [in the direction of the applied force]. Validate the Solution The led ha a negative acceleration, which indicate that the led i lowing down. In thee ort of copetition, ditance within the arena are all (le than 100 ), o acceleration ut be high to enure object top relatively quickly. For coparion, note that a vehicle which accelerate fro ret to 100 k/h (27.8 /) in 6.0 econd ha an acceleration of 4.6 / 2. Thu, the agnitude of thi reult ee reaonable. 10. Solution i iilar to Practice Proble 8 and 9. (a) The force of friction of the ice on the curling tone i 0.249 N. (b) The coefficient of kinetic friction between the ice and the curling tone i 1.27 10 3. Fa θ F N y Fg Ff x 11. Conceptualize the Proble - Begin by drawing a free-body diagra. - Motion i in the horizontal direction. Becaue the velocity i contant, the acceleration i zero in thi direction. - The x-coponent of the applied force and the frictional force act in oppoite direction. - The y-coponent of the applied force and the gravitational force act in the oppoite direction to the noral force. - Becaue the coefficient of kinetic friction i not given, the proble ut be olved fro the force alone. The frictional force, F f, acting on the cart Identify the Variable Known Iplied Unknown F = 95 N a = 0 / 2 F f v 1 = 1.2 / θ = 35 Develop a Strategy Calculation To find the frictional force, apply F co θ F f = a = 0 Newton econd law in the x- and F co θ = F f y-direction, noting that the acceleration i zero in each direction. 89
Subtitute and olve. F f = (95 N)(co 35 ) F f = 77.82 N F f 78 N The frictional force i 78 N. Note there i no need to apply Newton law in the y-direction becaue the proble i already olved. Validate the Solution The frictional force i about 2/3 of the applied force, which ee reaonable conidering the angle of the applied force. 12. Solution i iilar to Practice Proble 8. (a) The horizontal force i 58 N. (b) The kateboard accelerate at 16 / 2, in the direction of the applied force. 13. Solution i iilar to Practice Proble 8 and 9. The bicycle travel 6.7 after the brake are applied. Practice Proble Solution Student Textbook page 173 174 14. Frae the Proble - Draw a free body diagra repreenting the force acting on the wier. The net force i the u of the force acting on the wier. - Ue both a cale diagra and a atheatical olution to deterine the net force acting on the wier. Find the reultant force acting on the wier uing a cale diagra and a atheatical olution Variable and Contant Known Iplied Unknown F 1 = 35.0 N[N] F reultant F 2 = 20 N[E] θ reultant Strategie Scale Vector Diagra Method Draw a cale vector diagra, adding the vector tip to tail. Meaure the length of the reultant vector. Ue a cale factor to deterine the agnitude of the force. Ue a protractor to eaure the angle. F net = 4.1 c F net = (4.1 c) ( 10 N ) c F net = 41 N θ = [N30 E] 90
Uing the cale diagra ethod the net force i 41 N[N30 E]. Scale 1 c : 10 N N+ F 2 = 20.0 N[E] F 1 = 35.0 N[N] F net = 41 N[N30 E] 30 E+ Coponent Method Draw each vector with it tail at the origin at an x-y-coordinate yte where +y coincide with north and +x coincide with eat. Find the angle with the nearet x-axi. North coincide with the y-axi o the angle i 90. Eat coincide with the x-axi o the angle i 0. Find the x-coponent of each force vector. F 1x = F 1 co 90 = (35 N)(0.0) = 0.0 N F 2x = F 2 co 0 = (20 N)(1.0) = 20.0 N Find the y-coponent of each force vector. F 1y = F 1 in 90 = (35 N)(1.0) = 35.0 N F 2y = F 2 in 0 = (20 N)(1.0) = 0.0 N Make a table in which to lit the x-and y-coponent. Add the to find the coponent of the reultant vector. Vector x-coponent y-coponent F 1 0.0 N 35.0 N F 2 20.0 N 0.0 N F net 20.0 N 35.0 N Scale 1 c : 10 N N+ F 1y = 35.0 N F 2x = 20.0 N E+ 91
Ue the Pythagorean Theore to find the agnitude of the net force. Ue trigonoetry to find the angle θ. The angle will be in the firt quadrant. F net 2 = (Fx net ) 2 + (F y net ) 2 F net 2 = (20 N) 2 + (35.0 N) 2 F net 2 = 400 N 2 + 1225 N 2 F net 2 = 1625 N 2 F net = 40.31 N F net = 40 N tan θ = 35.0 N 20 N tanθ = 1.75 θ = tan 1 1.75 θ = 60 The net force acting on the wier i 40[N60 N] or 40[N30 E]. Validate The agnitude and direction of the reultant force vector are correct. Uing the cale diagra and the coponent ethod give nearly the ae reult. The agnitude of the net force i 41 N for the cale diagra and 40 N for the coponent ethod. The direction are both [N30 E]. 15. Frae the Proble - Connect the force acting on each object on a cale vector diagra. The reultant repreent the u of the force acting on each object. Find the reultant force acting on each object pictured Variable and Contant Known Iplied Unknown F 1 F 2 F reultant F 3 F 4 θ reultant Strategy and Calculation Draw a cale vector diagra, adding the vector tip to tail. Meaure the length of the reultant vector. Ue a cale factor to deterine the agnitude of the force. Ue a protractor to eaure the angle. (a) F net = 0.86c F net = (0.86 c) ( 50 N ) c F net = 43 N θ = [E] The reultant force i 43 N[E]. (b) F net = 0.148 c F net = (0.148 c) ( 50 N ) c F net = 7.4 N θ = [N] 92
The reultant force i 7.4 N[N]. (c) F net = 0.3 c F net = (0.3 c) ( 50 N ) c F net = 15 N θ = [E] The reultant force i 1 N[E]. (d) F net = 0.3 c F net = (0.3 c) ( 50 N ) c F net = 15 N θ = [W28 S] The reultant force i 15 N[W28 S]. Validate The agnitude and direction of the reultant force vector are reaonable. 16. Frae the Proble - Sketch the pulling force vector. - Reolve the pulling force vector into it x- and y-coponent. y Sketch of pulling force F (a) Find the coponent of the pulling force in the direction of travel (x-coponent). (c) Find the coponent of the pulling force perpendicular to the direction of travel (y-coponent). Variable and Contant Known Iplied Unknown F = 1500 N[N] Fx θ = 15 = 1500 N 15 F x = 1.4 10 3 N F y = 3.9 10 2 N x Strategy and Calculation Find the x-coponent and y-coponent of the force vector. F x = F co 15 = (1500 N)(0.9659) = 1448.9 N F x = 1.4 10 3 N F y = F co 15 = (1500 N)(0.2588) = 388.2 N F y = 3.9 10 2 N Validate The agnitude and direction of the reultant force vector are reaonable. F y 93
17. Frae the Proble -Draw a free body diagra of the lawn ower howing all the force acting on it. -Reolve the applied force into it vertical and horizontal coponent. - The tudent puhing down on the lawn ower will increae the noral force, o that the force of gravity on the lawn ower ay be aller than the noral force. -Newton econd law applie to thi proble. (a) Find the vertical and horizontal coponent of the applied force (b) Calculate the noral force upporting the lawn ower while it i being puhed (c) Calculate the net force propelling the ower if a frictional force of 85 N exit (d) Calculate the horizontal acceleration of the lawn ower Free body diagra of lawn ower F ay = 86 N F a = 150 N 35 F ax = 120 N F g = 245 N F N = 331 N F f = 85 N Variable and Contant Known Iplied Unknown F a = 150 N g = 9.81 F 2 ax a θ = 35 F ay = 25 kg F g F N Strategy and Calculation Find the vertical and horizontal coponent of the applied force. Aue the lawn ower i oving to the right, which i poitive and let up be poitive. F ax = F a co 35 = (150 N)(0.8192) = 122.88 N F ax = 120 N F ay = F a in 35 = (150 N)(0.5736) = 86.03 N F ay = 86 N The horizontal coponent of the applied force i 120 N to the right. The vertical coponent of the applied force i 86 N down. Calculate the noral force upporting the lawn ower. It i equal to all the downward force acting on it, but oppoite in ign. F g = g = 25 kg ( 9.81 N kg ) F g = 245.25 N F N = F g + F ay = 245.25 N + 86 N F N = 331.25 N F N = 3.3 10 2 N Calculate the net force on the lawn ower if a 85 N friction force exit. The u of the vertical force i zero, and the acceleration i in the direction of the u of the horizontal force only. 94
F net = F x + F f F net = 1.22.88 N 85 N = 37.88 N F net = 38 N The net force on the lawn ower i 38 N to the right along the horizontal. Calculate the horizontal acceleration of the lawn ower. a = F net a = 38 N 25 kg = 1.52 2 a = 1.5 2 The lawn ower horizontal acceleration i 1.5 / 2 to the right. Validate The agnitude and unit of the coponent force, noral force, net force and acceleration ee reaonable. Practice Proble Solution Student Textbook page 182 18. Conceptualize the Proble - Repreent direction to the right a poitive and to the left a negative. - The two car accelerate a though they were one object, with a a equal to the total a of both vehicle. - Only part of the applied force i tranitted by the rope to the econd vehicle. - By Newton third law, the force applied to the rope (the tenion it ut tand without breaking) i equal to the force the rope applie to the econd vehicle. (a) The total force, F, needed to accelerate the vehicle (b) The tenion force, F t, that the rope ut withtand Identify the Variable Known Iplied Unknown 1 = 1700 kg F t = F 2 a 2 = 2400 kg F v 1 = 0 F 2 v 2 = 15 k h t = 11 Develop a Strategy Apply the definition of acceleration and change to tandard unit. Calculation a = v t a = (15 0) k h 1000 k 11 1h 3600 a = 0.3787 2 Ue Newton econd law to find the agnitude of the force neceary to caue thi acceleration. F = a F = (1700 kg + 2400 kg) ( 0.3787 ) 2 F = 1553 kg 2 F 1.6 10 3 N 95
(a) The total force neceary to accelerate the two vehicle i about 1.6 10 3 N. Ue Newton econd law to find the agnitude of the force needed to caue the acceleration of the econd vehicle. F 2 = a F 2 = 2400 kg 0.3787 2 F 2 = 908.88 N Find the agnitude of the reaction F t = F 2 force applied by the econd vehicle F t = 908.88 N to the rope. F t 9.1 10 2 N (b) The rope ut be trong enough to withtand a tenion force of about 9.1 10 2 N. Validate the Solution The rope applie a force of 9.1 10 2 N to the firt car, which oppoe the force cauing it to accelerate. The agnitude of the acceleration of the firt car can be calculated fro thi force. F = a a = F 1 a = (1553 908.88) N 1700 kg a = 0.379 2 The acceleration calculated earlier for the two car together wa 0.379 / 2, in perfect agreeent (within 3 ignificant figure) with that calculated for the firt car here. 19. Solution i iilar to Practice Proble 18. (a) The total force neceary to accelerate the train i about 21 N[to the right]. (b) The econd child feel a tenion of about 15 N[to the right]. 20. Solution i iilar to Practice Proble 9 and 18. (a) The frictional force acting on the toboggan i about 74 N. (b) The tenion in the rope attached to the econd toboggan i about 34 N. Practice Proble Solution Student Textbook page 186 21. Conceptualize the Proble - Begin fraing the proble by drawing a free-body diagra of the force acting on the peron on the cale. - The force acting on the peron are gravity ( F g ), which i downward, and the noral force of the cale, which i upward. - According to Newton third law, when the peron exert a force ( F PS ) on the cale, it exert an equal and oppoite force ( F SP ) on the peron. Therefore, the reading on the cale i the ae a the force that the peron exert on the cale. - The otion i in one direction o let up be poitive and down be negative. - Becaue the otion i in one dienion, perfor calculation with agnitude only. - The elevator i accelerating upward and the peron i accelerating at the ae rate a the elevator. However, the elevator i lowing down, o, even though the acceleration i in the poitive direction, the ign of the acceleration i negative. 96
- Apply Newton econd law to find the agnitude of ( F SP ). The ign of the acceleration, and the reading on the cale, ( F SP ) Identify the Variable Known Iplied Unknown = 64 kg g = 9.81 / 2 F SP a = 0.59 / 2 F PS F g Develop a Strategy Calculation Before applying Newton econd law to the force, note that ince the elevator i accelerating in the poitive direction, but it i lowing down, the ign of the acceleration ut be negative. Apply Newton econd law F SP + F g = a to the free-body diagra. F SP = F g + a F SP = ( g) + a Subtitute and olve. F SP = (64 kg)(9.81 / 2 ) + (64 kg)( 0.59 / 2 ) F SP = 590.1 N F SP 5.9 10 2 N The cale ha a reading of about 5.9 10 2 N[up]. Validate the Solution Becaue of the negative ign of the acceleration (i.e. directed downward), the reading on the cale i lightly le than it would be if the elevator had zero acceleration. In that cae, the cale would read (64 kg)(9.81 / 2 ) = 628 N. Thi agree with the experience of riding in an elevator: travelling upward, a it coe to a top, you feel lighter. 22. Solution i iilar to Practice Proble 21. The cale ha a reading of about 6.9 10 2 N[up]. 23. Solution i iilar to Practice Proble 21. The cale ha a reading of about 5.9 10 2 N[up]. Practice Proble Solution Student Textbook page 191 24. Conceptualize the Proble - Begin fraing the proble by drawing a free-body diagra. - Define a coordinate yte with the x-axi parallel to the hill and the y-axi perpendicular to the hill. On the coordinate yte, draw the force and coponent of force acting on the car. - Let the direction pointing down the hill be the poitive direction. - The coponent of gravity parallel to the hill caue the car to accelerate down the hill. 97
F N - A friction i neglected, it i not neceary to find the noral force, which i perpendicular to the hill. - The acceleration of the car down the hill i found by applying Newton econd law to the force or coponent of force parallel to the hill. - Fro the acceleration, the final velocity can be found fro the appropriate kineatic equation. Fg Proble Tip Conidering force in direction parallel or perpendicular to the direction of otion enable one to ue agnitude (intead of vector) in the calculation. +y F N The final peed, v 2, of the car at the botto of the hill +x F g θ F g Identify the Variable Known Iplied Unknown = 1975 kg g = 9.81 / 2 F a a v 2 d = 42 v 1 = 0 / F g F g θ = 15 Fg Develop a Strategy Apply Newton econd law to the force parallel to the hill and olve for the acceleration parallel to the hill. Subtitute and olve. The acceleration down the rap i about 2.539 / 2. Apply an appropriate kineatic relation, involving diplaceent, acceleration and final velocity. Subtitute and olve. The car i travelling 15 / at the botto of the hill. Calculation F net = a F g = a g in θ = a a = g in θ a = (9.81 / 2 )(in 15 ) a = 2.539 / 2 v 2 2 = v2 1 + 2ad, where a = a v 2 2 = 0 + 2(2.539 /2 )(42 ) v 2 = 14.6 / v 2 15 / Validate the Solution The acceleration of the car i le than 9.81 / 2, a expected. The final peed, 15 /, i equivalent to 54 k/h, which ee reaonable for a car rolling 42 down a hill. 25. Solution i iilar to Practice Proble 24. (a) Ignoring friction, the acceleration down the rap i about 1.2 / 2. (b) Conidering friction, the acceleration of the bicycle down the rap i about 0.16 / 2. (c) The cyclit take 12 to get to the botto of the rap. 26. Conceptualize the Proble - Begin fraing the proble by drawing a free-body diagra. (The free-body diagra for thi proble i the ae a that in the Saple Proble, Sliding Down an Inclined Plane, in the Student Text). 98
- Define a coordinate yte with the x-axi parallel to the lope and the y-axi perpendicular to the lope. On the coordinate yte, draw the force and coponent of force acting on the kier. - Let the direction pointing down the lope be the poitive direction. - The coponent of gravity parallel to the lope caue the kier to accelerate down the lope. - Friction between the kier and the lope oppoe the otion. To find the noral force that i needed to deterine the agnitude of the frictional force, apply Newton econd law to the force or coponent of force that are perpendicular to the lope. - The acceleration perpendicular to the lope i zero. - The acceleration of the kier down the lope i found by applying Newton econd law to the force or coponent of force parallel to the lope, but note that in thi cae the peed i contant, which iplie the acceleration i zero. The coefficient of kinetic friction, µ k, between the kier and the lope Identify the Variable Known Iplied Unknown θ = 3.5 g = 9.81 / 2 F FN F f a = 0 / 2 F g F g a = 0 / 2 µ k Develop a Strategy Apply Newton econd law to the force perpendicular to the rap to find the frictional force. The acceleration i zero in the perpendicular direction. A the a i unknown, do not ubtitute and olve. Apply Newton econd law to the force parallel to the rap, including the force of friction. Note that the a cancel. Subtitute and olve. The coefficient of kinetic friction between the kier and the lope i 0.061. Calculation F net = a F N + F g = a = 0 F N + ( g co θ) = 0 F N = g co θ F net = a F f + F g = a µf N + g in θ = a = 0 µ(g co θ) + g in θ = 0 µ = in θ co θ µ = 0.061 Validate the Solution The coefficient of kinetic friction i only lightly greater than zero. Coparion with Table 4.5 how that the coefficient of kinetic friction for ice on ice i 0.03, o a value of 0.061 for wood or fibergla on now ee reaonable. 99
Practice Proble Solution Student Textbook page 194 27. Solution i iilar to Practice Proble 26. The coin travel 0.34 up the board before topping. 28. Solution i iilar to Practice Proble 26. The coefficient of friction between the crate and rap i about 0.37. Practice Proble Solution Student Textbook page 197 29. Conceptualize the Proble - In each cae, the a i oving; therefore, it ha oentu. - The direction of an object oentu i the ae a the direction of it velocity. The oentu, p, of the given object Identify the Variable Known (a) b = 0.250 kg (b) t = 7.5 10 6 kg (c) j = 4.00 10 5 kg (d) e = 9.11 10 31 kg v b = 46.1 /[E] v t = 125 k/h[w] v j = 755 k/h[s] v e = 6.45 10 6 /[N] Unknown pb pt pj pe Develop a Strategy Ue the equation that define oentu. Calculation pb = b vb pb = (0.250 kg)(46.1 /[E]) pb = 11.525 kg /[E] pb 11.5 kg /[E] (a) The oentu of the baeball i about 11.5 kg /[E]. pt = t vt pt = (7.5 10 6 kg)(125 k h pt = 2.6042 10 8 kg /[W] pt 2.6 10 8 kg /[W] 1000 1k 1h 3600 [W]) (b) The oentu of the train i about 2.6 10 8 kg /[W]. pj = j vj pj = (4.00 10 5 kg)(755 k h pt = 8.3889 10 7 kg /[S] 1000 1k 1h 3600 [S]) pt 8.39 10 7 kg /[S] (c) The oentu of the jet i about 8.39 10 7 kg /[S]. 100
pe = e ve pe = (9.11 10 31 kg)(6.45 10 6 /[N]) pe = 5.8760 10 24 kg /[N] pe 5.88 10 24 kg /[N] (d) The oentu of the electron i about 5.88 10 24 kg /[N]. Validate the Solution A expected, the high a object like the train and jet have uch higher oenta than low a object like the baeball or very low a object like the electron. Though the electron i travelling at a very high peed copared to the other object, it a i o low that it oentu will reain aller than the other object. Practice Proble Solution Student Textbook page 200 30. Conceptualize the Proble - The ledgehaer exert an average force on the pike for a period of tie. The product of thee quantitie i defined a ipule - Ipule i a vector quantity. - The direction of the ipule i the ae a the direction of it average force. The ipule, J, of the interaction Identify the Variable Known F = 2125 N[down] t = 0.0205 Develop a Strategy Ue the equation that define ipule. Unknown J Calculation J = F t J = (2125 N[down])(0.0205 ) The ledgehaer trike the pike with an ipule of 43.6 N [down]. Validate the Solution Round the value in the data to 2000 N[down] and 0.02 and do ental ultiplication. The product i 40.0 N [down]. The anwer, 43.6 N [down], i very cloe to thi etiate. 31. Conceptualize the Proble - The ipule i the product of the average force exerted by the car on the wall and the tie interval. - Ipule i a vector quantity. - The direction of the ipule i the ae a the direction of it average force. The ipule, J, of the interaction J = 43.562 N [down] J 43.6 N [down] 101
Identify the Variable Known F = 1.23 10 7 N[S] t = 21.0 / = 21.0 10 3 Develop a Strategy Ue the equation that define ipule. Unknown J Calculation J = F t J = (1.23 10 7 N[S])(21.0 10 3 ) The car hit the wall with an ipule of 2.58 10 5 N [S]. Validate the Solution If the tie i rounded to 20 10 3, then the ultiplication can eaily be checked: the reult i 2.46 10 5 N [S], cloe to the calculated value of 2.58 10 5 N [S]. 32. Conceptualize the Proble - The ipule i the product of the average force exerted by the car on the wall and the tie interval. - Ipule i a vector quantity. - The direction of the ipule i the ae a the direction of it average force. The average force, F, of the interaction J = 2.583 10 5 N [S] J 2.58 10 5 N [S] Identify the Variable Known J = 2.58 10 5 N [S] t = 57.1 / = 57.1 10 3 Develop a Strategy Ue the equation that define ipule and olve for the average force. Calculation J = F t F = J t F = 2.58 10 5 N [S] 57.1 10 3 F = 4.5184 10 6 N[S] Unknown F F 4.52 10 6 N[S] The car trike the wall with an average force of 4.52 10 6 N[S]. Validate the Solution In coparion with proble 3, the ae ipule wa applied over a longer tie interval, an effect that i deigned to reduce the average force. The reultant force i lower here than in proble 3, a expected. 102
Practice Proble Solution Student Textbook page 203 33. Conceptualize the Proble - The a and velocitie before and after the interaction are known, o it i poible to calculate the oentu before and after the interaction. - Moentu i a vector quantity, o all calculation ut include direction. - Let the initial direction of the ball be forward. - You can find the ipule fro the change in oentu. The ipule, J, of the racquet on the ball Identify the Variable Known Iplied Unknown = 0.060 kg v 1 = 0 /[forward] J v 2 = 43 /[forward] Develop a Strategy Ue the ipule-oentu theore to calculate the ipule. Calculation F t = v2 v 1 F t = (0.060 kg)(43 /)[forward] (0.060 kg)(0 /)[forward] F t = 2.58 kg /[forward] The ipule of the racquet on the ball i about 2.6 kg /[forward]. Validate the Solution Check the unit: kg / = (kg / 2 ) () = N 34. Conceptualize the Proble - The a and velocitie before and after the interaction are known, o it i poible to calculate the oentu before and after the interaction. - Moentu i a vector quantity, o all calculation ut include direction. - Since the otion i all in one dienion (i.e. the horizontal dienion), ue plu and inu to denote direction. Let the initial direction be the poitive direction. - You can find the ipule fro the change in oentu. The ipule, J, of the bat on the ball F t 2.6 kg /[forward] Identify the Variable Known = 0.35 kg v 1 = 46 / v 2 = 62 / Develop a Strategy Ue the ipule-oentu theore to calculate the ipule. Unknown J Calculation F t = v2 v 1 F t = (0.35 kg)( 62 /) (0.35 kg)(46 /) F t = 37.8 kg / F t 38 kg / 103
The ipule of the bat on the ball i about 38 kg /, and the negative ign indicate the direction i away fro the batter. Validate the Solution The ipule i in the oppoite direction to the ball initial velocity, o it i expected that the ipule will be negative, and it i. 35. Conceptualize the Proble - The book i diplaced through a known ditance fro ret, o it velocity before triking the floor can be calculated fro the appropriate kineatic equation. - After triking the floor with a certain velocity, the floor exert an upward force on the book over a certain length of tie and becaue of thi ipule, the book final velocity will be zero. - The a and velocitie before and after the interaction can thu be deterined, o it i poible to calculate the oentu before and after the interaction, and the ipule. - Moentu i a vector quantity, o all calculation ut include direction. - Since the otion i all in one dienion (i.e. the vertical dienion), ue plu and inu to denote direction (with the agnitude of the quantitie). Let the downward direction be the negative direction. - You can find the ipule fro the change in oentu. The ipule, J, of the floor on the book Identify the Variable Known Iplied Unknown = 1.5 kg v 1 = 0 / J y = 1.75 a = 9.81 / 2 v 2 Develop a Strategy Apply the kineatic equation that relate acceleration, diplaceent and initial velocity to final velocity. Calculation v2 2 = v2 1 + 2a y v2 2 = (0 /)2 + 2( 9.81 / 2 )( 1.75 ) v2 2 = 34.335 2 / 2 v2 2 =±5.860 / = 5.860 / v 2 2 Chooe the negative ign becaue the book i travelling in the negative direction before hitting the floor. Ue the ipule-oentu theore to calculate the ipule. Note that a the book trike the floor, it initial velocity i v 1 = 5.860 / and it final velocity, after interaction with the floor i v 2 = 0 /. F t = v 2 v 1 F t = (1.5 kg)(0 /) (1.5 kg)( 5.860 /) F t = 8.789 kg / F t 8.8 kg / The floor exert an ipule on the book of about 8.8 kg / (in the upward direction). Validate the Solution The ipule of the floor on the book i expected to be in a direction oppoite to the book downward velocity, that i upward, and it i. 104
Chapter 5 Review Anwer to Proble for Undertanding Student Textbook page 208 209 23. (a) 2500 N 3200 N 6000 N F net = 3200 N 2500 N + 6000 N F net =+300 N (b) v v Fnet a = v + 300 N a = 400 kg v a =+ 075. v a + 08. 2 2 24. The acceleration i due to the net force in the x direction. The u of the force cancel in the y direction, o only force in the x direction need be calculated. F g = 15 kg 9.81 2 F g = 147.15 N F x = F co θ F x = 45 N co 40 F x = (45 N)(0.7660) F x = 34.47 N F f = 28 N F net = F x F f F net = 34.47 N 28 N F net = 6.47 N a = F net a = 6.47 N 15 kg a = 0.43 2 105
25. (a) Add the u of all the vertical force to obtain the noral force. N + F N = 637.02 N F f = 382.21 N F x = 389.71 N 30 + E F g = 412.02 N F y = 225 N F = 450 N F g = 42 kg 9.81 2 F g = 412.02 N F y = F in θ F y = 450 N in 30 F y = 450 N (0.5) F y = 225 N F N = F g + F y F N = 412.02 N + 225 N F N = 637.02 N F f = µ k F N F f = (0.60)(637.02 N) F f = 382.21 N F f = 3.80 3.8 10 10 2 N 2 N (b) The acceleration of the grocery cart i in the horizontal direction. Find the u of all the force in the x direction. (The u of the force in the y direction i zero.) F x = F co θ F x = 450 N co 30 F x = (450 N)(0.8660) F x = 389.71 N F net = F f + F x F net = 382.21 N + 389.71 N F net = 7.5 N 106
a = F net a = 7.5 N 42 kg a = 0.179 2 a = 0.18 2 1 c = 20 N F 2 = 60 N[N] F 3 = 30 N[N35 ] F 1 = 40 N[S] N + + E 26. Refer to the cale diagra below: 1 c = 20 N F 3 F net = 50 N[E70 N] θ = [E70 ] F 2 N + F 1 + E F 1 = 40 N[S] F 2 = 60 N[N] F 3 = 30 N[N35 E] F net = (2.5 c)(20 N F net = 50 N θ = [E70 N] F net = 50 N[E70 N] c ) 27. (a) Refer to the diagra and calculation below: 40 kg F = 48 N 20 kg F 40 on 20 20 kg 107
T = 40 kg + 20 kg = 60 kg a = F net = 48 N T 60 kg = 0.8 [ ] 2 b) F 40 on 20 = a = ( 20 kg ) ( 0.8 2 ) [ ] F 40 on 20 = 16 N[ ] 28. (a) The initial velocity of the watch in a frae attached to the elevator i 0.0 / and the initial acceleration i 9.81 / 2, where poitive i up and negative i down. v = 00. and a = 98. 2 i (b) The initial velocity of the watch in a frae attached to the building i 3.5 / and the initial acceleration i 9.81 / 2, where poitive i up and negative i down. v = 35. and a = 98. 2 i 29. (a) Acceleration of the crate i 1.35 / 2, where the direction of the applied force i poitive. a = Fnet Fapp + F a = f 425N + ( µg) a = 425 N (0.500)(68.0 kg)(9.81 ) a = 2 68.0 kg a 1.35 2 (b) To ove the crate at contant velocity require that the net force be zero. The applied force ut have a agnitude of 334 N. Fnet = 0 Fapp + F f = 0 Fapp = F f Fapp = ( µg) ) Fapp = (0.500)(68.0 kg) (9.81 2 Fapp 334 N 30. The agnitude of the force acting on the ball i 1.2 N. F = a F = (0.24 kg) ( 5.0 2 ) F = 1.2 N 108
31. (a) Acceleration of the brick i (b) The ratio of the applied force to 0.063 / 2. the a i 0.40 N/kg. d = v i t + 1 2 a t 2 F = 4.0 N 10.0 kg d = 0.0 + 1 2 a t 2 F a = 2 d = 0.40 N kg t 2 a = 2(2.0 ) (8.0 )2 a 0.063 2 (c) The nuber do not agree becaue, fro Newton econd law, acceleration i the ratio of net force over a, not applied force over a a found in part (b). Friction ut be included to deterine the net force. If the acceleration i 0.063 / 2, the force of friction ut be Fnet = a Fapp + F f = a F f = F app + a F f = 4.0N + (10.0 kg) ( 0.0625 ) 2 F f 3.4 N 32. The car went 11.4 before coing to a top. Firt find the car acceleration while braking. Let the direction of the car otion be poitive. Then, ue the acceleration to find the topping ditance of the car. Fnet = a vf 2 = vi 2 + 2a d a = Ff a = µ g a = (0.70) ( 9.81 2 ) a = 6.867 2 d = v2 f vi 2 2a ( ) 2 [( 0 45 k h k d = ) 2 ( 6.867 2 d d = 11.4. 4 d 11 )( 1000 )( )] h 2 3600 33. (a) The woan velocity jut before touching the floor wa 5.4 / downward. (Note that the anwer could have been poitive or negative. Since you know that the woan wa oving down, you can chooe the negative direction for her velocity.) vf 2 = vi 2 + 2a d v f = vi 2 + 2a d (0 ) 2 ( v f = + 2 9.81 ) 2 ( 1.5) 2 v f ±5.4 109
(b) The floor exerted a force of 3.7 10 4 N up on the woan feet. F = a F = v f v i t F = 55.0 kg (0 ) ( 5.4 ) 8.00 10 3 F 3.7 10 4 N[up] 34. The new acceleration i one half of the original aount, ince the new a i twice the original aount. 35. (a) The new acceleration will be 5/2 tie the original acceleration. F 1 = 1 a1 F2 = 2 a2 2 = 2 1 a2 = 5 F 1 a1 = F 1 a2 = F 2 F2 = 5 2 1 F 1 a2 = 5 1 2 2 a 1 (b) The body will travel a ditance 3 4 a 1 t 2 greater than the ditance travelled by the original body. If the bodie tarted fro ret, the econd would have travelled 5 or 2.5 tie a far a the original travelled. 2 ( ) d 1 = v i t + 1 2 a 1 t 2 d 2 = 0 + 1 5 2 2 a 1 t 2 d 1 0 + 1 d 2 = v i t + 1 2 a 2 t 2 2 a 1 t 2 d ( ) 2 = 5 d 2 = v i t + 1 5 2 2 a 1 t 2 d 1 2 d 2 = v i t + 5 4 a 1 t 2 d 2 = v i t + 1 2 a 1 t 2 + 3 4 a 1 t 2 d 2 = d 1 + 3 4 a 1 t 2 36. (a) The agnitude of the net force acting horizontally on the box i 9.00 N. Fnet = a a = v f v i t F net = v f v i t F net = 45.0 kg 1.50 1.00 2.50 F net = 9.00 N (c) In order to find the frictional force in part (b), it i neceary to find the horizontal (x) coponent of the applied force. Therefore, olve part (c) firt. The horizontal coponent of the applied force i 141 N. F x(app) = F app co θ F x(app) = (205 N)(co 46.5 ) F x(app) 141 N 110
(b) The frictional force acting on the box i 132 N in the direction oppoite to the applied force. F x(net) = F x(app) + F x(f) F x(f) = F x(net) F x(app) F x(f) = 9.0 N 141 N F x(f) 132 N (d) The coefficient of friction i 0.450. Firt find the noral force. The noral force i entirely in the y direction, a i the gravitational force. F N + F g + F y(app) = 0 F f = µ k F N F N = F g F y(app) F N = ( g) F app in θ F N = (45.0 kg) ( 9.81 ) (205 N)(in 46.5 ) 2 F N 293 N µ k = F F F N µ k = 132 N 293 N µ k 0.450 37. Moentu of the bowling ball v v p = v v p = ( 50. kg)( 35. [ N] ) v kg p = 17. 5 [ N] v kg p 18 [ N] 38. Ma of the car p = v = p v = 4.2 104 kg [W] 28 [W ] = 1.5 10 3 kg 39. Velocity of the in-line kater 40. Velocity of the golf ball p = v g vg = 6 v6 p v = 66.0 kg [S] v = 55.0 kg 1.2 = 120. [S] vg = b v b g vg = (5.0 kg)(6.0 [forward]) 5.0 10 3 kg vg 6.0 10 3 [forward] 41. (a) Ipule for a peron knocking on (b) Ipule for a wooden allet triking a door a gong J = F t J = F t J = (9.1 N[E])(2.5 10 3 ) J = (4.2 N[S])(8.6 10 3 ) J 2.3 10 3 N [E] J 3.6 10 2 N [S] 111
42. Average applied force 43. Duration of the contact J = F t J = F t F J = J t = t F F 8.8 kg = 2.3 10 3 F t = 2.0 N 55 N 3.8 10 3 N t 3.6 10 2 44. (a) Ipule of the hockey puck laphot F t = p F t = vf v i ( F t = 0.300 kg 9.2 [S]) 0.300 kg ( 44 [N]) kg F t = 2.76 [S] 13.2 kg [N] kg F t = 2.76 [S] + 13.2 kg [S] kg F t = 15.96 F t 16 kg [S] [S] (b) Duration of the interaction kg F t = 15.96 [S] kg 15.96 [S] t = F kg 15.96 [S] t = 2.5 10 3 N[S] t 6.4 10 3 45. Velocity of the bullet jut before it hit the wood F t = p F = v f v i t 0 (2.5 kg)(3.5 F = [W]) 3.5 10 4 F 2.5 10 4 N[W] F 2.5 10 4 N[E] 46. Force that the eat belt exert on the crah duy on ipact v v v v vi + vf F t = vf vi d = t 2 v v v ( f vi) F = 2 v t t = d vi + vf v 75. 0 kg 0 28 [forward] F = 2 0. 714 t = v 4 + F = 294. 10 N[backward] [forward] [forward] 28 0 10. v 4 t = 0. 0714 F 29. 10 N[backward] ( ) 112