Answer Key
4 b. Two chains connected to the load each Chapter can withstand a tension of 15,000 N. 1. You and your bike have a combined mass Can the load be safely lifted at 1.5 m/s 2? ϭ ϩ of 80 kg. How much braking force has to Fload Flift Fovercome gravity be applied to slow you from a velocity of ϭ ma ϩ mg 5 m/s to a complete stop in 2 s? ϭ (1.5 ϫ10 3 kg)(1.5 m/s 2) ϩ v Ϫ v Ϫ 3 2 a ϭ ᎏf ᎏi ϭ ᎏᎏᎏ0.0 m/s 5.0 m/s (1.5 ϫ10 kg)(9.80 m/s ) t Ϫ t 2.0 s Ϫ 0.0 s f i ϭ 2.25 ϫ10 3 N ϩ 1.47 ϫ10 4 N ϭ 2.5 m/s 2 ϭ 1.7 ϫ10 4 N F ϭ ma The load can be safely lifted ϭ ϫ Ϫ 2 80 kg ( 2.5 m/s ) because the total force on the ϭ Ϫ 200 N chains is less than their combined 4 2. Before opening his parachute, a sky diver capability of 3.0 ϫ10 N with a mass of 90.0 kg experiences an 4. In a lab experiment, you attach a 2.0-kg upward force from air resistance of 150 N. weight to a spring scale. You lift the scale a. What net force is acting on the sky diver? and weight with a constant reading of ϭ 22.5 N. Fgravity mg a. What is the value and direction of the ϭ 90.0 kg ϫ 9.80 m/s 2 acceleration on the weight? ϭ 882 N downward ϭ Ϫ Fnet Fscale Fgravity F ϭ F ϩ F net air resistance gravity ϭ 22.5 N Ϫ (2.0 kg)(9.80 m/s 2) F ϭ 150 N ϩ (Ϫ882 N) net ϭ 2.9 N upward ϭ Ϫ 732 N ϭ 730 N downward a ϭ ᎏmᎏ Fnet b. What is the sky diver’s acceleration? 2.0 kg F ϭ ᎏᎏ a ϭ ᎏnetᎏ 2.9 N m ϭ 0.69 m/s 2 Ϫ730 N ϭ ᎏᎏ Ϫ 90 .0 kg ϭ 6.9 ϫ10 1 m/s 2 McGraw-Hill McGraw-Hill Companies, Inc. ϭ Ϫ 8.1 m/s 2 b. How far do you lift the weight in the ϭ 8.1 m/s 2 downward first 2.0-s interval? d ϭ d ϩ v t ϩ ᎏ1ᎏat 2 3. A large helicopter is used to lift a heat f i i 2 pump to the roof of a new building. The ϭ 0 m ϩ (0 m/s)(2.0 s) ϩ mass of the helicopter is 5.0 ϫ10 3 kg and ᎏ1ᎏ(6.9 ϫ10 Ϫ1 m/s 2)(2.0 s) 2 the mass of the heat pump is 1500 kg. 2 a. How much force must the air exert on ϭ 1.4 m the helicopter to lift the heat pump with an acceleration of 1.5 m/s 2? 5. As a large jet flies at a constant altitude, its engines produce a forward thrust of Copyright Copyright © Glencoe/McGraw-Hill, a division of The F ϭ F ϩ F net lift overcome gravity 8.4 ϫ10 5 N. The mass of the plane is ϭ ma ϩ mg 2.6 ϫ10 5 kg. ϭ (6.5 ϫ10 3 kg)(1.5 m/s 2) ϩ (6.5 ϫ10 3 kg )(9.80 m/s 2) ϭ 7.3 ϫ10 4 N upward
Physics: Principles and Problems Supplemental Problems Answer Key 75 Answer Key
Chapter 4 continued a. What is the forward acceleration of the b. How long does it take for the 2.0-kg plane, ignoring air resistance? mass to fall to the floor? ϭ F ma Solve d ϭ d ϩ v t ϩ ᎏ1ᎏat 2 for t f i i 2 a ϭ ᎏFᎏ ϭ ϭ m while di 0 m, vi 0 m/s ϫ 5 ϭ ᎏ8.4 1ᎏ0 N t ϭ ᎏ2dᎏ 2.6ϫ105 kg Ίa ϭ 2 (2)(1 .5 m) 3.2 m/s ϭ ᎏᎏ Ί2 b. How much upward force must the air 4.2 m/s exert on the plane when it is flying ϭ 0.85 s horizontally? 7. A man is standing on a scale inside an Because the plane is not changing airplane. When the airplane is traveling altitude, horizontally (in other words, the vertical ϭ ϭ Ϫ Fnet 0, so Flift Fgravity acceleration of the plane is zero) the ϭ Ϫ scale reads 705.6 N. What is the vertical Flift Fgravity acceleration of the plane in each of the ϭ Ϫ ϫ 5 Ϫ 2 (2.6 10 kg)( 9.80 m/s ) following situations? ϭ 2.5 ϫ10 6 N a. When the scale reads 950.0 N. ϭ 6. Two masses are tied to a rope on a pulley, Fg mg as shown below. F m ϭ ᎏg g F ϭ ma Frictionless pulley net ϭ ϩ Ϫ Fnet Fscale ( Fg) Copyright © Glencoe/McGraw-Hill, a division of The of division a Glencoe/McGraw-Hill, © Copyright ma ϭ F Ϫ F 2.0 kg 0.8 kg scale g F Ϫ F a ϭ ᎏᎏscale g m 1.5 m F Ϫ F ϭ ᎏᎏscale g F ᎏᎏg a. When the system is released from this g position, what is the acceleration of the g(F Ϫ F ) ϭ ᎏᎏscale g 2.00-kg mass? Fg
F ϭ F Ϫ F ϩ (9.80 m/s 2)(950.0 N Ϫ 705.6 N) Inc.Companies,McGraw-Hill net g-large g-small ϭ ᎏᎏᎏᎏ Ϫ mlarge g msmall g 705.6 N ϭ (2.00 kg)(9.80 m/s 2) Ϫ ϭ 3.39 m/s 2 (0.80 kg)(9.80 m/s 2) b. When the scale reads 500.0 N. ϭ ϭ 11.8 N downward Fg mg F F a ϭ ᎏᎏ m ϭ ᎏᎏg m g ϭ ᎏ11.8 N F ϭ ma 2.8 kg net F ϭ F ϩ (ϪF ) ϭ 4.2 m/s 2 downward net scale g
76 Supplemental Problems Answer Key Physics: Principles and Problems Answer Key
Chapter 4 continued ϭ Ϫ ma Fscale Fg a. What is the average acceleration on the 2 Ϫ ball as it was struck (in m/s )? Fscale Fg a ϭ ᎏᎏ v Ϫ v m ϭ ᎏf ᎏi ϭ a , where vi 0 km/h Ϫ t Fscale Fg ϭ ᎏᎏ 100 0 m 1 h F (1.8 ϫ10 2 km/h) ᎏᎏ ᎏᎏ ᎏᎏg 1 km 36 00 s g a ϭ ᎏᎏᎏ (0.50 ms) ᎏ1 ᎏs g(F Ϫ F ) 100 0 ms ϭ ᎏsca leᎏg Fg ϭ ᎏ50 mᎏ/s 5.0 ϫ10Ϫ4 s (9.80 m/s 2)(500.0 N Ϫ 705.6 N) ϭ ᎏᎏᎏᎏ ϭ 1.0 ϫ10 5 m/s 2 705.6 N b. What is the force exerted on the club? ϭ Ϫ 2.86 m/s 2 F ϭ ma 8. An airboat glides across the surface of the 1 kg ϭ (45 g) ᎏᎏ ϫ (1.0 ϫ10 5 m/s 2) water on a cushion of air. Perform the 100 0 g following calculations for a boat in which ϭ 4.5 ϫ10 3 N the mass of the boat and passengers is 450 kg. c. What is the force exerted on the club by a. If there is no friction, how much force the ball? ϭ Ϫ must the propeller fan exert on the air Fball Fclub to accelerate the boat at 5.00 m/s 2? ϭ Ϫ 4.5 ϫ10 3 N F ϭ ma 10. A package of instruments is attached to a 2 ϭ (450 kg)(5.00 m/s ) helium-filled weather balloon that exerts an ϭ 2.2 ϫ10 3 N upward force of 45 N. b. If the actual acceleration with the fan a. If the instrument package weighs 10.0 kg, generating the force calculated in part a will the balloon be able to lift it? 2 ϭ is only 4.95 m/s , how much friction Flift ma does the air cushion exert on the boat? F a ϭ ᎏliftᎏ ϭ Ϫ m afriction aideal aactual
McGraw-Hill McGraw-Hill Companies, Inc. ϭ 45 N Ffriction ma friction ϭ ᎏᎏ Ϫ 10 .0 kg m(aideal aactual )(450 kg) ϭ 4.5 m/s 2 ϭ (5.00 m/s 2 Ϫ 4.95 m/s 2) The balloon cannot lift the package ϭ 22 N because the upward acceleration is c. What is the upward force exerted by the less than the downward acceleration air cushion on the boat? of gravity, 9.80 m/s 2. ϭ Ϫ ϭ Ϫ Flift Fgravity mg Alternative calculation: ϭ Ϫ Ϫ 2 ϭ 2 (450 kg)( 9.80 m/s ) Fgravity (10.0 kg)(9.80 m/s ) 3 ϭ Copyright Copyright © Glencoe/McGraw-Hill, a division of The ϭ 4.4 ϫ10 N 98.0 N, which exceeds the upward force 9. A golf ball with a mass of 45 g is struck by a club, leaving the tee with a speed of b. What is the upward acceleration if the 1.8 ϫ10 2 km/h. The period of acceleration instruments weigh 2.0 kg? ϭ Ϫ was 0.50 m/s. Fnet Flift Fgravity ϭ Ϫ Fnet ma mg Physics: Principles and Problems Supplemental Problems Answer Key 77 Answer Key
Chapter 4 continued ϭ 45 N Ϫ (2.0 kg)(9.80 m/s 2) 14. A 0.100-kg weight is attached to a spring ϭ 25.4 N scale. Find the reading on the scale for each of the following situations. F a ϭ ᎏnetᎏ a. When the scale is not moving. m ϭ Fg mg ϭ ᎏ25.4 ᎏN 2.0 kg ϭ (0.100 kg)(9.80 m/s 2) ϭ 13 m/s 2 upward ϭ 0.980 N 11. A 12-kg block sits on a table. A 10.0-kg b. When the scale accelerates at 2.4 m/s 2 block sits on top of the 12-kg block. If there in the horizontal direction. is nothing on top of the 10.0-kg block, The vertical component of the what is the force that the table exerts on acceleration is zero, so the reading the 12-kg block? on the scale is calculated the same ϭ ϩ Ftable Fblock 1 Fblock 2 way as when the scale is not moving. ϭ ϭ (12 kg)(9.80 m/s 2) ϩ Fg mg (10.0 kg)(9.80 m/s 2) ϭ (0.100 kg)(9.80 m/s 2) ϭ 220 N ϭ 0.980 N 12. A box experiences a net force of 41 N while 15. A penny is dropped from the top of a it is being lifted. What is the acceleration of 30.0-m-tall tower. The tower, however, is the box? not located on Earth. The penny has a mass ϭ Fnet ma of 2.5 g and experiences a gravitational F force of 0.028 N. a ϭ ᎏnet a. What is acceleration due to gravity on m The of division a Glencoe/McGraw-Hill, © Copyright this planet? ϭ ᎏ41 N 9.7 kg F ϭ ᎏg ag ϭ 4.2 m/s 2 m 0.028 N 3 ϭ ᎏᎏ 13. A rope can withstand 1.000 ϫ10 N of 0.0025 kg tension. If the rope is being used to pull a ϭ 11 m/s 2 10.0-kg package across a frictionless surface, what is the greatest acceleration that will b. After 1.00 s, the penny has a velocity of not break the rope? 10.1 m/s. Assuming the force exerted on ϭ the penny by air resistance is uniform
F ma Inc.Companies,McGraw-Hill and independent of speed, what is the a ϭ ᎏFᎏ magnitude of the force of air resistance m on the penny? ϫ 3 ϭ ᎏ1.0 00 ᎏ10 N ϭ 10.0 kg Fnet maෆ 2 2 ϭ ϩ Ϫ ϭ 1.00 ϫ10 m/s Fg ( Fa) ⌬ F ϭ F Ϫ ma , where a ϭ a ϭ ᎏᎏv a g ෆ ⌬t ⌬ ϭ F Ϫ m ᎏᎏv g ⌬t
78 Supplemental Problems Answer Key Physics: Principles and Problems Answer Key
Chapter 4 continued v Ϫ v b. A force of 150.0 N is applied to the ϭ F Ϫ m ᎏf ᎏi g ⌬t sled and produces an acceleration of 3.00 m/s 2. What is the magnitude of ϭ Ϫ 0.028 N (0.0025 kg) the force of friction that resists the Ϫ ᎏᎏᎏ10.1 m/s 0.0 m/s acceleration? 1.00 s ϭ Fnet ma ϭ 0.0028 N F ϭ F Ϫ F The air resistance is in the opposite net applied f ϭ Ϫ direction from the force of gravity and ma Fapplied Ff the penny’s motion, as it should be. ϭ Ϫ Ff Fapplied ma 16. The combined mass of a sled and its rider is ϭ 150.0 N Ϫ (46.4 kg)(3.00 ms 2) 46.4 kg. The sled is pulled across a frozen lake so that the force of friction between the ϭ 10.8 N sled and the ice is very small. a. Assuming that friction between the sled and the ice is negligible, what force is required to accelerate the sled at 3.45 m/s 2? F ϭ ma ϭ (46.4 kg)(3.45 m/s 2) ϭ 1.60 ϫ10 2 N McGraw-Hill McGraw-Hill Companies, Inc. Copyright Copyright © Glencoe/McGraw-Hill, a division of The
Physics: Principles and Problems Supplemental Problems Answer Key 79 Answer Key
Chapter 5 1. A small plane takes off and flies 12.0 km in a direction southeast of the airport. At this point, following the instructions of an air traffic controller, the plane turns 20.0° to the east of its original flight path and flies 21.0 km. What is the magnitude of the plane’s resultant displacement from the airport?
N
E
A R
B
R2 ϭ A2 ϩ B2 Ϫ 2AB cos
R ϭ ͙(12.0ෆkm)ෆ2 ϩෆ(21.0 ෆkm) 2ෆϪ 2(1 2.0ෆ km )(21.0ෆෆkm)(c ෆos 16 0.0°)ෆ ϭ 32.5 km 2. A hammer slides down a roof that makes a 32.0° angle with the horizontal. What are the magnitudes of the components of the hammer’s velocity at the edge of the roof if it is moving at a speed of 6.25 m/s?
y Copyright © Glencoe/McGraw-Hill, a division of The of division a Glencoe/McGraw-Hill, © Copyright
v 32.0 Њ x x v vy 58.0 Њ
Ͼ Ͻ Fourth quadrant: vx 0 and vy 0. ϭ vx v cos ϭ (6.25 m/s)(cos Ϫ32.0°)
ϭ 5.30 m/s Inc.Companies,McGraw-Hill ϭ vy v sin ϭ (6.25 m/s)(sin Ϫ32.0°) ϭ Ϫ 3.31 m/s
80 Supplemental Problems Answer Key Physics: Principles and Problems Answer Key
Chapter 5 continued 3. A worker has to move a 17.0-kg crate along a flat floor in a warehouse. The coefficient of kinetic friction between the crate and the floor is 0.214. The worker pulls horizontally on a rope attached to the crate, with a 49.0-N force. What is the resultant acceleration of the crate? y-direction: ϭ ϭ FN Fg mg x-direction: ϭ Ϫ Fnet , x Fp Ff ϭ ϭ ma x ma ϭ ϭ Ff kFN kmg ϭ Ϫ ma Fp kmg F Ϫ mg a ϭ ᎏp ᎏk m Ϫ 2 ϭ ᎏᎏᎏᎏᎏ49.0 N (0.214)(17.0 kg)(9.80 m/s ) 17.0 kg ϭ 0.785 m/s 2
4. To get a cart to move, two farmers pull on ropes attached to the cart, as shown below. One farmer pulls with a force of 50.0 N in a direction 35.0° east of north, while the other exerts a force of 30.0 N in a direction 25.0° west of north. What are the magnitude and the direction of the combined force exerted on the cart?
N
25.0 Њ 35.0 Њ ϭ ϩ ϭ ϩ E Rx Ax Bx A cos 1 B cos 2
McGraw-Hill McGraw-Hill Companies, Inc. ϭ ϩ ϭ ϩ Ry Ay By A sin 1 B sin 2 ϭ ͙ෆ2 ϩෆ2 R Rx Ry ϭ ͙ෆෆϩෆෆ2 ϩෆෆϩ ෆෆ2 (A co s 1 B cos 2) (A sin 1 B sin 2) ϭ ͙((50.0ෆෆN)(co sෆ 55.0 °)ෆϩ (3 ෆ0.0 N) (cosෆ 1 15.0°)ෆ)ෆ2 ϩ ((50.0ෆN)(sinෆෆ55.0° )ෆ ϩ (3 0.0ෆ N) (sinෆ 1 15.0°)ෆ)ෆ2 ϭ 70.0 N
R A sin ϩ B sin ϭ Ϫ1 ᎏᎏy ϭ Ϫ1 ᎏᎏᎏ1 2 tan tan ϩ Rx A cos 1 B cos 2 Copyright Copyright © Glencoe/McGraw-Hill, a division of The ϩ ϭ Ϫ1 ᎏᎏᎏᎏᎏ(50.0 N)(sin 55.0°) (30.0 N)(sin 115.0°) tan (50.0 N)(cos 55.0°) ϩ (30.0 N)(cos 115.0°) ϭ 76.8° R ϭ 70.0 N at 76.8° north of east
Physics: Principles and Problems Supplemental Problems Answer Key 81 Answer Key
Chapter 5 continued 5. Takashi trains for a race by rowing his canoe Ϫ ϭ x-direction: Fp Ff ma on a lake. He starts by rowing along a straight a ϭ 0 (since v ϭ constant) path. Then he turns and rows 260.0 m west. F Ϫ F ϭ 0 If he then finds he is located 360.0 m exactly p f ϭ north of his starting point, what was his Fp Ff displacement along the straight path? ϭ ϭ kFN kFg N F B ϭ ᎏᎏp k Fg
R ϭ 45.0ᎏ N A 235 N ϭ 0.191 E 7. A rod supports a 2.35-kg lamp, as shown below.
ϭ ϭ Rx 0.0 m, Ry 360.0 m ϭ Ϫ ϭ Bx 260.0 m, By 0.0 m 25.0 Њ R ϭ A ϩ B, A ϭ R Ϫ B ϭ Ϫ ϭ Ϫ Ϫ Ax Rx Bx 0.0 m ( 260.0 m) ϭ 260.0 m a. What is the magnitude of the tension in ϭ Ϫ ϭ Ϫ the rod? Ay Ry By 360.0 m 0.0 m ϭ 360.0 m y-direction: The of division a Glencoe/McGraw-Hill, © Copyright ϭ ͙ෆ2 ϩෆ2 T ϭ F A Ax Ay y g ϭ ϭ ͙(260.0ෆෆm) 2 ϩෆ(360. 0ෆ m) 2 T sin mg mg ϭ 444.1 m T ϭ ᎏᎏ sin Ϫ Ay ϭ tan 1 ᎏᎏ (2.35 kg)(9.80 m/s 2) A ϭ ᎏᎏᎏ x sin 25.0° ϭ tan Ϫ1 ᎏ360.ᎏ0 m ϭ 54.16° ϭ 54.5 N 260.0 m
A ϭ 444.1 m at 54.16° north of east b. Calculate the components of the force Inc.Companies,McGraw-Hill that the bracket exerts on the rod. 6. Mira received a 235-N sled for her birthday. x-direction: She takes the sled out to a flat snowy field. Ϫ ϭ When she pushes it with a 45.0-N horizon- Tx Fx 0 tal force, it slides along at a constant speed. F ϭ T cos What is the coefficient of kinetic friction x between the sled and the snow? ϭ (54.5 N)(cos 25.0°) ϭ ϭ y-direction: FN Fg Fx 49.4 N, inward
82 Supplemental Problems Answer Key Physics: Principles and Problems Answer Key
Chapter 5 continued y-direction: ϭ Fy Ty ϭ mg ϭ (2.35 kg)(9.80 m/s 2) ϭ Fy 23.0 N, upward
8. A 25.0-kg crate has an adjustable F ϭ 70.0 N handle so that it can be pushed or at 330.0 ° pulled by the handle at various angles. F ϭ 70.0 N at 0.0 ° Determine the acceleration of the crate for each situation shown in the diagram, given that the coefficient of sliding friction between the floor and the bottom of the crate is 0.20. m ϭ 25.0 kg (a) (b) For (a) F ϭ 70.0 N y-direction: at 30.0 ° ϭ ϭ FN Fg mg ϭ FN mg x-direction: ϭ Ϫ Fnet , x Fp Ff (c) ϭ ϭ Fnet , x ma x ma ϭ Ϫ ma Fp Ff ϭ ϭ Ff kFN kmg ϭ Ϫ ma Fp kmg F Ϫ mg F a ϭ ᎏp ᎏk ϭ ᎏᎏp Ϫ g m m k McGraw-Hill McGraw-Hill Companies, Inc. ϭ ᎏ70.0ᎏN Ϫ (0.20)(9.80 m/s 2) 25.0 kg ϭ 0.84 m/s 2
For (b) y-direction: ϭ ϩ FN Fg Fy ϭ ϩ FN mg F sin x-direction: Copyright Copyright © Glencoe/McGraw-Hill, a division of The ϭ Ϫ Fnet , x Fp Ff ϭ ϭ Fnet , x ma x ma ϭ Ϫ ma Fp Ff
Physics: Principles and Problems Supplemental Problems Answer Key 83 Answer Key
Chapter 5 continued ϭ Fp F cos ϭ ϭ ϩ Ff kFN k(mg F sin ) ϭ Ϫ ϩ ϭ Fnet , x F cos k(mg F sin ) ma F a ϭ ᎏnetᎏ m F cos Ϫ (mg ϩ F sin ) ϭ ᎏᎏᎏᎏk m F cos Ϫ F sin ϭᎏᎏᎏk Ϫ g m k (70.0 N)(cos 30.0°) Ϫ (0.20)(70.0 N)(sin 30.0°) ϭ ᎏᎏᎏᎏᎏᎏ Ϫ (0.20)(9.80 m/s 2) 25.0 kg ϭ 0.18 m/s 2
For (c) y-direction: ϭ Ϫ FN Fg Fy ϭ Ϫ FN mg F sin x-direction: ϭ Ϫ Fnet , x Fp Ff ϭ ϭ Fnet , x ma x ma ϭ Ϫ ma Fp Ff ϭ The of division a Glencoe/McGraw-Hill, © Copyright Fp F cos ϭ ϭ Ϫ Ff kFN k(mg F sin ) ϭ Ϫ Ϫ ϭ Fnet , x F cos k(mg F sin ) ma F a ϭ ᎏnetᎏ m F cos Ϫ (mg Ϫ F sin ) ϭ ᎏᎏᎏᎏk m F cos ϩ F sin ϭᎏᎏᎏk Ϫ g m k McGraw-Hill Companies, Inc.Companies,McGraw-Hill ϩ ϭ ᎏᎏᎏᎏᎏᎏ(70.0 N)(cos 30.0°) (0.20)(70.0 N)(sin 30.0°) Ϫ (0.20)(9.80 m/s 2) 25.0 kg ϭ 0.74 m/s 2
84 Supplemental Problems Answer Key Physics: Principles and Problems Answer Key
Chapter 5 continued 9. A child shoves a small toboggan weighing 100.0 N up a snowy hill, giving the toboggan an initial speed of 6.0 m/s. If the hill is inclined at an angle of 32° above the horizontal, how far along the hill will the toboggan slide? Assume that the coefficient of sliding friction between the toboggan and the snow is 0.15. y-direction: ϭ ϭ Fnet, y ma y 0 Ϫ ϭ FN Fgy 0 ϭ ϭ FN Fgy mg cos
x-direction: ϭ ϭ Fnet , x ma x ma ϭ Fgx mg sin Ϫ ϭ Fgx Ff ma ϭ ϭ Ff kFN kmg cos Ϫ ϭ mg sin kmg cos ma ϭ Ϫ a g(sin k cos )
2 ϭ 2 ϩ Ϫ vf vi 2a(df di) ϭ 2 ϩ 0 vi 2ad v 2 v 2 ϭ Ϫ ᎏiᎏ ϭ ᎏᎏᎏi d Ϫ 2a 2g(sin k cos ) 2 ϭ ᎏᎏᎏᎏᎏ(6.0 m/s ) (2)(9.80 m/s 2)(sin 32° Ϫ (0.15)cos 32°) ϭ 4.6 m, up the hill
McGraw-Hill McGraw-Hill Companies, Inc. 10. Two objects are connected by a string passing over a frictionless, massless pulley. As shown below, the block is on an inclined plane and the ball is hanging over the top edge of the plane. The block has a mass of 60.0 kg, and the coefficient of kinetic friction between the block and the inclined plane is 0.22. If the block moves at a constant speed down the incline, and the ball rises at a constant speed, what is the mass of the hanging ball?
m1 Copyright Copyright © Glencoe/McGraw-Hill, a division of The
m2 35.0 Њ
Physics: Principles and Problems Supplemental Problems Answer Key 85 Answer Key
Chapter 5 continued v is constant, so there is no acceleration.
Ball ( m2): y-direction: ϭ ϭ Fnet, y m2ay 0 Ϫ ϭ FT Fg 0 ϭ FT m2g
Block ( m1): y-direction: (parallel to the incline) ϭ ϭ Fnet, y m1ay 0 ϭ ϭ FN Fgy 0 ϭ ϭ FN Fgy m1g cos ϭ ϭ Ff kFN km1g cos
x-direction: (perpendicular to the incline) ϭ ϭ Fnet , x m1ax 0 Ϫ Ϫ ϭ Fgx Ff FT 0 ϭ Fgx m1g sin Ϫ Ϫ ϭ m1g sin km1g cos FT 0 m g sin Ϫ m g cos Ϫ m g ϭ 0 1 k 1 2 The of division a Glencoe/McGraw-Hill, © Copyright m g sin Ϫ m g cos m ϭ ᎏᎏᎏᎏ1 k 1 2 g ϭ Ϫ m1 sin km1 cos ϭ (60.0 kg)(sin 35.0°) Ϫ (0.22)(60.0 kg)(cos 35.0°) ϭ 24 kg McGraw-Hill Companies, Inc.Companies,McGraw-Hill
86 Supplemental Problems Answer Key Physics: Principles and Problems