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# Practice Exam 2 for General Physics 1 PHYS113

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This 18 page Study Guide was uploaded by Ericah Notetaker on Tuesday March 29, 2016. The Study Guide belongs to PHYS113 at Kansas State University taught by Horton Smith, Glenn A. in Fall 2015. Since its upload, it has received 17 views. For similar materials see General Physics 1 in Art at Kansas State University.

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Date Created: 03/29/16

Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: General Physics Exam 2 Practice, version 1 1. A jogger jogs 57 m east, then 39 m north. How far is the jogger from where she or he started? A. 2.2×10 +03 m , calculated from (57 m)*(39 m) B. 96 m, calculated from (57 m)+(39 m) C. 69 m, calculated from sqrt((57 m) +(39 m) ) D. 0.68, calculated from (39 m)/(57 m) E. 18 m, calculated from (57 m)-(39 m) 2. What direction is the jogger from where he/she started? A. due east B. There is not enough information given to solve this problem. C. 34° north of east D. due north E. 34° east of north 3. A truck travels 83 km/hr up a 24 ° slope. What is the truck's vertical velocity? A. 0 km/hr B. 76 km/hr C. -34 km/hr D. 34 km/hr E. -83 km/hr page 1 of 7 A stone is thrown straight upward from the top of a cliff. It leaves the thrower's hand at A, reaches its highest point at B, and lands at D. 4. If the speed of the stone at point A is 14.5 m/s and its speed at point D is 57.5 m/s, how high above the top of the cliff does it go? (The acceleration of gravity is 9.8 m/s down.) A. 11 s, calculated from (14.5 m/s) /(2*(9.8 m/s ))2 2 B. 1.5 s, calculated from (14.5 m/s)/(9.8 m/s ) C. 85 m, calculated from (14.5 m/s)*(57.5 m/s)/(9.8 m/s ) 2 2 D. 1.5 m, calculated from (14.5 m/s)/(9.8 m/s ) E. 11 m, calculated from (14.5 m/s) /(2*(9.8 m/s ))2 5. How much time does it take the stone to go from A to D? 2 A. 4.39 s, calculated from ((57.5 m/s)-(14.5 m/s))/(9.8 m/s ) B. 158 s, calculated from ((57.5 m/s) -(14.5 m/s) )/(2*(9.8 m/s ))2 2 2 2 C. 158 m, calculated from ((57.5 m/s) -(14.5 m/s) )/(2*(9.8 m/s )) D. 7.35 s, calculated from (14.5 m/s)/(9.8 m/s ) + (57.5 m/s)/(9.8 m/s )2 6. How high is the cliff? A. 158 m, calculated from ((57.5 m/s) -(14.5 m/s) )/(2*(9.8 m/s )) 2 B. 7.35 m, calculated from (14.5 m/s)/(9.8 m/s ) + (57.5 m/s)/(9.8 m/s ) 2 2 2 2 C. 158 s, calculated from ((57.5 m/s) -(14.5 m/s) )/(2*(9.8 m/s )) D. 4.39 m, calculated from ((57.5 m/s)-(14.5 m/s))/(9.8 m/s ) 2 page 2 of 7 7. What is the average velocity of the stone over the whole trip between points A and D? A. 72 m/s down B. 36 m/s down C. 36 m/s up D. 21.5 m/s up E. 21.5 m/s down 8. An "angry bird" is launched by a slingshot. The bird's initial velocity has a horizontal component 16 m/s and a vertical component 13 m/s (up). For some reason, the bird keeps its wings folded, so it travels like a cannonball. What is the maximum height of the angry bird, relative to the slingshot? A. 13 m, calculated from (16 m/s) /(2*(9.8 m/s )) 2 2 B. 8.6 m, calculated from (13 m/s) /(2*(9.8 m/s )) C. 1.3 s, calculated from (13 m/s)/(9.8 m/s ) 2 2 D. 21 m/s, calculated from sqrt((16 m/s) +(13 m/s) ) E. 2.7 m, calculated from 2*(13 m/s)/(9.8 m/s ) 9. How long is the "angry bird" in the air? 2 2 A. 13 m, calculated from (16 m/s) /(2*(9.8 m/s )) B. 21 m/s, calculated from sqrt((16 m/s) +(13 m/s) )2 2 C. 2.7 s, calculated from 2*(13 m/s)/(9.8 m/s ) D. 1.3 s, calculated from (13 m/s)/(9.8 m/s ) 2 2 E. 8.6 m, calculated from (13 m/s) /(2*(9.8 m/s )) 10. How far does the "angry bird" travel horizontally before it hits the ground? Assume the ground is level, it was launched from ground level, and it doesn't hit anything until it hits the ground. A. 8.6 m B. 42 m C. 13 m D. 21 m/s E. 87 m 11. If an object has zero acceleration, then A. there is at most one force acting on it. B. this tells us nothing about the forces on the object or their sum. C. there can be any number of forces acting on it, as long as the sum of those forces is zero. D. there are no forces acting on it. page 3 of 7 12. If the sum of forces on an object is zero, then A. it gradually comes to rest. B. it moves in a constant direction at constant speed. C. its instantaneous velocity is zero. D. it has constant, non-zero acceleration. E. it moves in a circle at constant speed. 3 13. A double trailer combination on the highway (figure above) consists of two 23×10 kg trailers towed by a 12×10 kg "tractor truck". If the truck accelerates at 0.11 m/s over level ground, what is the forward force on the rear trailer? Ignore air friction. +03 A. 2.5×10 N B. 3.8×10 +03N +03 C. 1.3×10 N D. 6.4×10 +03N E. 5.1×10+03N 14. What is the force exerted on the first trailer by the tractor truck? +03 A. 5.1×10 N forward B. 1.3×10 +03N backward +03 C. 3.8×10 N forward D. 6.4×10 +03N forward +03 E. 2.5×10 N backward 15. The backward force of the tractor truck's wheels on the ground is A. 3.8×10 +03N +03 B. 2.5×10 N C. 5.1×10 +03N +03 D. 6.4×10 N E. 1.3×10+03N page 4 of 7 16. An object moves in a circle at constant speed. Which statement below is correct? A. The acceleration and velocity are perpendicular to each other, with the acceleration pointing away from the center of the circle. B. The velocity and acceleration point in the same direction. C. The acceleration is zero. D. The velocity and acceleration point opposite each other. E. The acceleration and velocity are perpendicular to each other, with the acceleration pointing in to the circle. 17. A grapefruit hanging from a tree weighs 9.72 N. The force of gravity exerted on the earth by the grapefruit is A. much greater than 9.72 N B. much less than 9.72 N C. exactly 0.00 N D. 9.72 N E. 4.86 N 18. The grapefruit falls from the tree. The force of gravity exerted on the earth by the grapefruit as it falls is A. 9.72 N B. much less than 9.72 N C. much greater than 9.72 N D. 4.86 N E. exactly 0.00 N page 5 of 7 19. A football quarterback carries the football towards the goal line at an initial speed of 2.1 m/s. The quarterback's feet apply a force of 612 N to the ground. An opposing player applies an opposing force on the quarterback such that the quarterback's acceleration is 0.87 m/s away from the goal line. If quarterback's mass is 140 kg, what is the magnitude of the force exerted by the opposing player? +02 2 A. 4.9×10 N, calculated from (612 N) - (140 kg)*((0.87 m/s )) B. 7.3×10 +02 N, calculated from (612 N) - (140 kg)*(-(0.87 m/s )) C. 6.1×10 +02 N, calculated from (612 N) +02 2 D. 1.2×10 N, calculated from (140 kg)*(0.87 m/s ) E. 2.9×10 +02 kg·m/s, calculated from (140 kg)*(2.1 m/s) 20. What is the farthest distance forward the quarterback goes before being stopped? 2 2 A. -2.5 m, calculated from (-(2.1 m/s)) /(2*(-(0.87 m/s ))) B. 2.1 m/s, calculated from (2.1 m/s) C. 0 m, calculated from 0 D. 2.4 s, calculated from (2.1 m/s)/(0.87 m/s ) 2 2 E. -5.1 m, calculated from (-(2.1 m/s)) /(-(0.87 m/s )) page 6 of 7 Written answer question [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] (i) A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 750 N, overcoming the force of friction between the losing player's feet and the grass. Draw a free-body diagram for the losing player and write the net force equation, using variables for any unknowns. (ii) The mass of the losing player plus equipment is 80.0 kg. The losing player is accelerating at 1.10 m/s² backward. What is the force of friction between the losing player’s feet and the grass? [The above problem is adapted from an end-of-chapter problem in our textbook.] The End page 7 of 7 Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, write your name on the next page, and also the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down -12 12 -11 3 -1 -2 pico 10-9 p tera 10 9 T Gravitational constant G = 6.674 × 10 5 kg 2 nano 10 -6 n giga 10 6 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×102J/K milli 10-3 m kilo 10 3 k Ideal gas constant R = 8.31 J/(mole·K) centi 10 -2 c hecto 10 2 h Avagadro's number NA= 6.02×1023 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ ⃗ ω= Δθ hypotenuse = opposite + adjacent ⃗= Δt Δt Δω t hy a= Δ⃗v α= Δt s po e n Δt o use ⃗ θ(t)=θ +ω t+ α1 t2 p θ ∑ F=ma ⃗ 0 0 2 const o adjacent 1 2 2 2 ⃗(t)=⃗0+v⃗0+ 2 ⃗const ω =ω +0αΔθ 2 2 ∑ τ=I α Impulses and collisions v =v0+2aΔ x Impulse: F Δ t=Δ ⃗ 1 2 KE rotationω 2 KE linear mv 2 Collision: ∑ ⃗pinitial ⃗final 2 W=τΔθ W=F dcosθ Fd Simple harmonic motion and waves p=mv L=I ω ⃗ ⃗ Period: T=2π inertia √ restoring Some kinds of potential energy Uniform gravity: PE Gmgy Spring: F=k x Pendulum: F=mg x/L Spring: PE = kx 2 Frequency: f =1/T S 2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr f nn(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Area of a sphere: 4πr Other definitions and relations Density: ρ=m/V Pressure: P=F/ A Circular or rotational motion v2 2 In a fluid: Δ P=ρ gΔ y v=rω, acentrip =rω Buoyancy: F Bρ fluidg r 2 τ=r Fsinθ, I= ∑ mr Gravity (sphere): F =G m 1 2 G r2 Static equilibrium Friction: F Fμ F N ax=a y0 → ∑ F x ∑ F y0 Ideal gas: PV=nRT=NkT 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Name: Date: General Physics Exam 2 Practice, version 2 1. A jogger jogs 65 m east, then 35 m north. How far is the jogger from where she or he started? A. 30 m, calculated from (65 m)-(35 m) +03 2 B. 2.3×10 m , calculated from (65 m)*(35 m) C. 1×10 +02m, calculated from (65 m)+(35 m) D. 74 m, calculated from sqrt((65 m) +(35 m) ) E. 0.54, calculated from (35 m)/(65 m) 2. What direction is the jogger from where he/she started? A. due north B. due east C. There is not enough information given to solve this problem. D. 28° east of north E. 28° north of east 3. A truck travels 98 km/hr up a 11 ° slope. What is the truck's vertical velocity? A. -19 km/hr B. 96 km/hr C. -98 km/hr D. 19 km/hr E. 0 km/hr page 1 of 7 A stone is thrown straight upward from the top of a cliff. It leaves the thrower's hand at A, reaches its highest point at B, and lands at D. 4. If the speed of the stone at point A is 14.9 m/s and its speed at point D is 65.5 m/s, how high above the top of the cliff does it go? (The acceleration of gravity is 9.8 m/s down.) A. 1×10 +02 m, calculated from (14.9 m/s)*(65.5 m/s)/(9.8 m/s )2 2 B. 1.5 s, calculated from (14.9 m/s)/(9.8 m/s ) C. 11 s, calculated from (14.9 m/s) /(2*(9.8 m/s )) 2 2 D. 11 m, calculated from (14.9 m/s) /(2*(9.8 m/s )) E. 1.5 m, calculated from (14.9 m/s)/(9.8 m/s ) 5. How much time does it take the stone to go from A to D? 2 2 2 A. 208 m, calculated from ((65.5 m/s) -(14.9 m/s) )/(2*(9.8 m/s )) B. 8.2 s, calculated from (14.9 m/s)/(9.8 m/s ) + (65.5 m/s)/(9.8 m/s ) 2 2 2 C. 208 s, calculated from ((65.5 m/s) -(14.9 m/s) )/(2*(9.8 m/s )) D. 5.16 s, calculated from ((65.5 m/s)-(14.9 m/s))/(9.8 m/s )2 6. How high is the cliff? A. 208 s, calculated from ((65.5 m/s) -(14.9 m/s) )/(2*(9.8 m/s ))2 B. 208 m, calculated from ((65.5 m/s) -(14.9 m/s) )/(2*(9.8 m/s )) 2 2 C. 5.16 m, calculated from ((65.5 m/s)-(14.9 m/s))/(9.8 m/s ) D. 8.2 m, calculated from (14.9 m/s)/(9.8 m/s ) + (65.5 m/s)/(9.8 m/s ) 2 page 2 of 7 7. What is the average velocity of the stone over the whole trip between points A and D? A. 25.3 m/s up B. 80.4 m/s down C. 40.2 m/s down D. 40.2 m/s up E. 25.3 m/s down 8. An "angry bird" is launched by a slingshot. The bird's initial velocity has a horizontal component 18 m/s and a vertical component 17 m/s (up). For some reason, the bird keeps its wings folded, so it travels like a cannonball. What is the maximum height of the angry bird, relative to the slingshot? A. 3.5 m, calculated from 2*(17 m/s)/(9.8 m/s ) 2 2 B. 17 m, calculated from (18 m/s) /(2*(9.8 m/s )) C. 25 m/s, calculated from sqrt((18 m/s) +(17 m/s) ) 2 2 D. 15 m, calculated from (17 m/s) /(2*(9.8 m/s )) E. 1.7 s, calculated from (17 m/s)/(9.8 m/s ) 9. How long is the "angry bird" in the air? 2 2 A. 25 m/s, calculated from sqrt((18 m/s) +(17 m/s) ) B. 17 m, calculated from (18 m/s) /(2*(9.8 m/s )) 2 C. 3.5 s, calculated from 2*(17 m/s)/(9.8 m/s ) D. 1.7 s, calculated from (17 m/s)/(9.8 m/s ) 2 2 E. 15 m, calculated from (17 m/s) /(2*(9.8 m/s )) 10. How far does the "angry bird" travel horizontally before it hits the ground? Assume the ground is level, it was launched from ground level, and it doesn't hit anything until it hits the ground. A. 1.3×10 +02 m B. 62 m C. 25 m/s D. 15 m E. 17 m 11. If an object has zero acceleration, then A. this tells us nothing about the forces on the object or their sum. B. there are no forces acting on it. C. there can be any number of forces acting on it, as long as the sum of those forces is zero. D. there is at most one force acting on it. page 3 of 7 12. If the sum of forces on an object is zero, then A. it moves in a circle at constant speed. B. it gradually comes to rest. C. it has constant, non-zero acceleration. D. its instantaneous velocity is zero. E. it moves in a constant direction at constant speed. 3 13. A double trailer combination on the highway (figure above) consists of two 23×10 kg trailers towed by a 12×10 kg "tractor truck". If the truck accelerates at 0.15 m/s over level ground, what is the forward force on the rear trailer? Ignore air friction. +03 A. 8.7×10 N B. 6.9×10 +03N +03 C. 3.4×10 N D. 1.8×10 +03N E. 5.2×10+03N 14. What is the force exerted on the first trailer by the tractor truck? +03 A. 8.7×10 N forward B. 5.2×10 +03N forward +03 C. 1.8×10 N backward D. 3.4×10 +03N backward +03 E. 6.9×10 N forward 15. The backward force of the tractor truck's wheels on the ground is A. 3.4×10 +03N +03 B. 8.7×10 N C. 6.9×10 +03N +03 D. 1.8×10 N E. 5.2×10+03N page 4 of 7 16. An object moves in a circle at constant speed. Which statement below is correct? A. The velocity and acceleration point opposite each other. B. The acceleration and velocity are perpendicular to each other, with the acceleration pointing away from the center of the circle. C. The acceleration is zero. D. The acceleration and velocity are perpendicular to each other, with the acceleration pointing in to the circle. E. The velocity and acceleration point in the same direction. 17. A grapefruit hanging from a tree weighs 9.72 N. The force of gravity exerted on the earth by the grapefruit is A. 4.86 N B. exactly 0.00 N C. much greater than 9.72 N D. 9.72 N E. much less than 9.72 N 18. The grapefruit falls from the tree. The force of gravity exerted on the earth by the grapefruit as it falls is A. 4.86 N B. much less than 9.72 N C. exactly 0.00 N D. 9.72 N E. much greater than 9.72 N page 5 of 7 19. A football quarterback carries the football towards the goal line at an initial speed of 2.4 m/s. The quarterback's feet apply a force of 675 N to the ground. An opposing player applies an opposing force on the quarterback such that the quarterback's acceleration is 0.93 m/s away from the goal line. If quarterback's mass is 125 kg, what is the magnitude of the force exerted by the opposing player? +02 2 A. 7.9×10 N, calculated from (675 N) - (125 kg)*(-(0.93 m/s )) B. 3×10 +02 kg·m/s, calculated from (125 kg)*(2.4 m/s) C. 6.8×10 +02 N, calculated from (675 N) +02 2 D. 1.2×10 N, calculated from (125 kg)*(0.93 m/s ) E. 5.6×10 +02 N, calculated from (675 N) - (125 kg)*((0.93 m/s )) 20. What is the farthest distance forward the quarterback goes before being stopped? A. 0 m, calculated from 0 B. 2.4 m/s, calculated from (2.4 m/s) 2 C. 2.6 s, calculated from (2.4 m/s)/(0.93 m/s ) D. -3.1 m, calculated from (-(2.4 m/s)) /(2*(-(0.93 m/s ))) 2 2 E. -6.2 m, calculated from (-(2.4 m/s)) /(-(0.93 m/s )) page 6 of 7 Written answer question [Note: your solution will be graded for the correctness of the work shown as well as on the final answer. Show and explain all work! Be sure to include proper units in all answers!] (i) A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 750 N, overcoming the force of friction between the losing player's feet and the grass. Draw a free-body diagram for the losing player and write the net force equation, using variables for any unknowns. (ii) The mass of the losing player plus equipment is 80.0 kg. The losing player is accelerating at 1.10 m/s² backward. What is the force of friction between the losing player’s feet and the grass? [The above problem is adapted from an end-of-chapter problem in our textbook.] The End page 7 of 7

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