Solutions to Homework Set #7 Phys2414 Fall 2005

Save this PDF as:

Size: px
Start display at page:

Transcription

1 Solution Set #7 Solutions to Homework Set #7 Phys244 Fall 2005 Note: The numbers in the boxes correspond to those that are generated by WebAssign. The numbers on your individual assignment will vary. Any calculated quantities that involve these variable numbers will be boxed as well.. GRR 6.P.002. A sled is dragged along a horizontal path at a constant speed of.5 m/s by a rope that is inclined at an angle of 30.0 with respect to the horizontal (the figure below). The total weight of the sled is 470 N. The tension in the rope is 230 N. How much work is done by the rope on the sled in a time interval of 5.0 s? The formula for calculating the work done on an object is W = F d cos θ where F is the magnitude of the force applied, d is the distance traveled, and cos θ is the cosine of the angle between the force vector and the displacement vector. In this case the displacement is horizontal and equal to the speed times the time interval given, i.e. d = vt The work done is then given by W = F (vt) cos θ W = ( 230 N) (.5 m/s) ( 5.0 s) cos (30.0 ) W =.49 kj 2. GRR 6.P.005. A barge of mass kg is pulled along the Erie Canal by two mules, walking along towpaths parallel to the canal on either side of it. The ropes harnessed to the mules make angles of 45 to the canal. Each mule is pulling on its rope with a force of.0 kn. How much work is done on the barge by both of these mules together as they pull the barge 30 m along the canal? The total work done is just the sum of the work done by each individual force. W total = i W i

2 Solution Set #7 2 The work done by a force is given by W F = F d cos θ So the work done by a single mule is given by W mule = (.0 kn) ( 30 m) cos (45 ) = 92 kn So the total work done is just the work done by each mule W total = W mule + W mule = 2W mule W total = 84 kn 3. GRR 6.P.02. A plane weighing 220 kn (25 tons) lands on an aircraft carrier. The plane is moving horizontally at 64 m/s ( 43 mi/h) when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 83 m. (a) How much work is done on the plane by the arresting cables? Work total work done is equal to the change in kinetic energy. Since the arresting cables are the only force that does work on the plane, then the work done by the arresting cables is the total work. So the work done by the arresting cables is equal to the change in kinetic energy. W = K K = K f K i = 2 mv2 f 2 mv2 i m = W/g K = 2 m(0)2 2 ( 220 kn/9.8 m/s2 ) ( 64 m/s) 2 W = MJ (b) What is the force (assumed constant) exerted on the plane by the cables? The work done by a force is given by: W = F d cos θ Solving for the force F we have the following formula. We plug in our previously calculated value for work, and because the cables pull on the plane in the opposite direction of the

3 Solution Set #7 3 plane s motion we use 80 for the angle. F = W d cos θ = MJ ( 83 m cos (80 ) F = 553 kn 4. GRR 6.P.02. Justin moves a desk 4.8 m across a level floor by pushing on it with a constant horizontal force of 30 N. (It slides for a negligibly small distance before coming to a stop when the force is removed.) Then, changing his mind, he moves it back to its starting point, again by pushing with a constant force of 30 N. (a) What is the change in the desk s gravitational potential energy during the round-trip? The change in gravitational potential energy is given by the following formula. Since the initial and final positions are the same (h f h i ) = 0. So the change in gravitational potential energy is. U g = mgh f mgh i = mg(h f h i ) = mg(0) = 0 (b) How much work has Justin done on the desk? The total work is the sum of all the individual works W total = i W i The work done on a single trip is then: W singletrip = F d cos θ = ( 30 N) ( 4.8 m) cos (0 ) W singletrip =.49 kj So the total work done is just the sum of the two single trips. W total = 2W Singletrip = 2.98 kj (c) If the work done by Justin is not equal to the change in gravitational potential energy of the desk, then where has the energy gone? Friction is a non-conservative force, so the mechanical energy is not conserved. The energy is dissipated in the form of heat

4 Solution Set # GRR 6.P.028. When a 0.20 kg mass is suspended from a vertically hanging spring, it stretches the spring from its original length of 3.0 cm to a total length of 7.0 cm. The spring with the same mass attached is then placed on a horizontal frictionless surface. The mass is pulled so that the spring stretches to a total length of 0.0 cm; then the mass is released and it oscillates back and forth (the figure below). What is the maximum speed of the mass as it oscillates? This problem has two parts, one where the spring is hanging from the ceiling and another where the spring is on a frictionless horizontal surface. We use the information of how the spring behaves as you hang is from a ceiling in to get the spring constant k. We use Newtons second law on the mass to do this. F = m a The only forces acting on the spring are the force of the spring, and weight. Solving the equation for k gives us: F spring mg = 0 k = mg x = (.2 kg) (9.8 m/s2 ) ( ) cm k = 49 N/m In the second part we use the conservation of energy to find the maximum velocity of the block. The conservation of energy states: K + U s = W NC Where K is the change in kinetic energy, U is the change in potential energy, and W NC is the work done by non-conservative forces. We then plug in K = /2mv 2 and U s = /2kx 2, and since there is no friction there is no work done by non-conservative forces or W NC is zero. 2 mv2 f 2 mv2 i + 2 kx2 f 2 mx2 i = 0 We know the initial velocity is zero, and solving for the final velocity we get. vf 2 = k m (x2 i x 2 f) k m (x2 i x 2 f )

5 Solution Set #7 5 So if we want the final velocity to be a maximum, then obviously x f must be zero. We can then plug in our value for the spring constant k we calculated, the mass, and the initial stretch of the spring (which would be the stretched length 0 cm minus the unstretched length 3 cm. k m (x i) 49 N/m ((0 3.0 ) cm) 0.2 kg. m/s 6. GRR 6.P.029. A cart starts from position 4 in the figure below with a velocity of 3 m/s to the left. Find the speed with which the cart reaches positions, 2, and 3. Neglect friction. We want to start with the conservation of energy. K + U g = W NC We then plug in K = /2mv 2 and U = mgh, and use the fact that there is no friction so W NC = 0. We can then solve for v f. 2 mv2 f 2 mv2 i + mgh f mgh i = 0 vi 2 + 2g(h i h f ) speed at position For these parts we just plug in the values of initial velocity v i, initial height h i and final height h f. ( 3 m/s) 2 + 2(9.8 m/s 2 ) (20 0) m 23.7 m/s speed at position 2 ( 3 m/s) 2 + 2(9.8 m/s 2 ) (20 5) m 6.3 m/s

6 Solution Set #7 6 speed at position 3 ( 3 m/s) 2 + 2(9.8 m/s 2 ) (20 0) m 9. m/s 7. GRR 6.P.034. An object slides down an inclined plane of angle 30.0 and of incline length 2.0 m. If the initial speed of the object is 6.0 m/s directed down the incline, what is the speed at the bottom? Neglect friction. We can do this problem like problem # 6. We start with the conservation of energy, plug in expressions for kinetic energy K and potential energy U. There is no friction so W NC = 0. We then solve for v f. K + U g = W NC 2 mv2 f 2 mv2 i + mgh f mgh i = 0 vi 2 + 2g(h i h f ) Next we have to realize that the change in height (h i h f ) is related to the distance the block travels down the incline d. h i h f = d sin θ = ( 2.0 m) sin ( 30 ) = m Finally we just plug in values. (6.0 m/s) 2 + 2(9.8 m/s 2 ) ( m) 7.46 m/s 8. GRR 6.P.039. A spring with k = 0.0 N/m is at the base of a frictionless 30.0 inclined plane. A 0.50 kg object is pressed against the spring, compressing it 0.2 m from its equilibrium position. The object is then released. If the object is not attached to the spring, how far up the incline does it travel before coming to rest and then sliding back down? (See the figure below.) We start with the conservation of energy equation, where for this problem we include both a gravitational potential energy term U g and a spring potential energy term U s. K + U g + U s = W NC

7 Solution Set #7 7 Without friction we know that W NC = 0. We also know that since our initial and final velocities are zero, then the total change in kinetic energy is zero. K = 2 mv2 f 2 mv2 i = 0 0 = 0 Plugging in expressions for U g and U s we get the following formula. mgh f mgh i + 2 kx2 f 2 kx2 i = 0 We then have to relate the change in height to the distance traveled up the incline d: h f h i = d sin θ We then just solve for the distance up the incline d. d = kx 2 i 2mg sin θ = ( 0 N/m)( 0.2 m) 2 2 (0.50 kg) (9.8 m/s 2 ) sin (30 ) d = m 9. GRR 6.P.044. A spring gun (k = 28 N/m) is used to shoot a 56 g ball horizontally. Initially the spring is compressed by 9 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 2 cm. What is the speed of the ball when it hits the ground,.4 m below the spring gun? We start with the conservation of energy. Then plug in the expressions for K, U g, and U s, and since there is no friction or air resistance we can set W NC = 0. K + U g + U s = W NC 2 mv2 f 2 mv2 i + mgh f mgh i + 2 kx2 f 2 kx2 i = 0 Since the ball starts from rest we set the initial velocity v i to zero, and solve for the final velocity. We can then plug in the given values for the change in height h i h f, the spring constant k, the mass m, and the initial and final values of the spring compressions x i and x f. 2 (9.8 [m/s2 )(.4 0) m + 2g(h i h f ) + k m (x2 i x 2 f ) 28 N/m.056 kg ((.9 m)2 (.2 m) 2 ) 6.9 m/s

8 Solution Set # GRR 6.P.057. The power output of a cyclist moving at a constant speed of 5.0 m/s on a level road is 0 W. (a) What is the force exerted on the cyclist and the bicycle by the air? Power is given by the formula P = F v, where P is power, F is the force, and v is the velocity. So using that formula and solving for F gives: P = F v F = P/v = 0 W/ 5.0 m/s F = 22 N (b) By bending low over the handlebars, the cyclist reduces the air resistance to 6 N. If she maintains a power output of 0 W, what will her speed be? Solving the power formula for velocity gives: P = F v v = P/F = 0 W/ 26 N F = 6.9 m/s. GRR 6.P.078. [299744] A 0.50 kg block, starting at rest, slides down a 30.0 incline with kinetic friction coefficient 0.28 (the figure below). After sliding 83 cm down the incline, it slides across a frictionless horizontal surface and encounters a spring (k = 33 N/m). (a) What is the maximum compression of the spring? We should use the conservation of energy to solve this problem, so we start out with the conservation of energy equation. K + U g + U s = W NC First we look at the change in kinetic energy K. Since the initial and final velocities are both zero, then the change in kinetic energy is also zero. K = 2 mv2 f 2 mv2 i = 0 0 = 0 We then plug in expressions for U g and U s, and since there is friction we replace W NC with the work done by friction W f. mgh f mgh i + 2 kx2 f 2 kx2 i = W f

9 Solution Set #7 9 But we need to figure out what the work done by friction is. The formula for work done by a force is: W f = F f d cos (80 ) But now we need to find the force of friction. We start by use the expression the expression for force of kinetic friction (kinetic since the block is moving). F f = µ k F N. Drawing a free body diagram, and setting the axis up along the plane we get for the sum of forces perpendicular to the plane. F N mg cos (30 ) = 0 So the force of friction is then given by the following formula, and the work done by friction is then similarly given by the next formula. F f = µ k mg cos (30 ) W f = ( 0.28 ) (0.50 kg) (9.8 m/s 2 ) cos (30 ) ( 83 cm) cos (80 ) = J Plugging this into our equation for conservation of energy, and setting x i = 0 and (h i h f ) = d cos(30 ), we get: 2 kx2 f = mgd sin (30 ) + W f = (0.50 kg) (9.8 m/s 2 ) ( 83 cm) sin (30 ) J x f = 2 kx2 f =.04 J 2 (.04 J)/( 33 N/m) x f = 25 cm (b) After the compression of part (a), the spring rebounds and shoots the block back up the incline. How far along the incline does the block travel before coming to rest? Again we want to use the conservation of energy to solve this problem. And again for the same reason as in part a we want to set K = 0. K + U g + U s = W NC K = 2 mv2 f 2 mv2 i = 0 0 = 0 So plugging in the usual expressions for U g and U s and setting W NC = W f we get: mgh f mgh i + 2 kx2 f 2 kx2 i = W f

10 Solution Set #7 0 Again we replace the change in height with the distance up the plane times the sine of the angle of the plane, and we use the formula for work done by a force. h f h i = d sin θ W f = F f d cos (80 ) And finally we solve for d the distance up the plane the block slides. d = 2 kx2 i mg(sin θ + µ k cos θ) =.04 J mgd sin θ + F f d = 2 kx2 i (0.50 kg) (9.8 m/s 2 ) (sin (30 ) cos (30 )) d = 28.5 cm 2. GRR 6.TB.042. [29697] A roller coaster car (mass = 988 kg including passengers) is about to roll down a track (the figure below ). The diameter of the circular loop is 20.0 m and the car starts out from rest 40.0 m above the lowest point of the track. Ignore friction and air resistance and assume g = 9.8 m/s2. (a) At what speed does the car reach the top of the loop? We start with the conservation of energy equation, and plug in the usual expressions for K and U g, and set W NC = 0 since there is no friction. K + U g = W NC 2 mv2 f 2 mv2 i + mgh f mgh i = 0 We then solve for the final velocity. vi 2 + 2g(h i h f ) (0) 2 + 2(9.8 m/s)(40 20) m 9.8 m/s (b) What is the force exerted on the car by the track at the top of the loop? The question ask about a force, so we start with a free body diagram, the free body diagram only has two forces: mg pointing straight down, and F N also pointing down since the track is above the cart at the top of the loop. Newton s Second law then states: F = m a

11 Solution Set #7 But we know that the acceleration is the centripetal acceleration, so a = v 2 /r. Solving for F N gives: F N = mv2 r mg = (988 kg) (9.8 m/s)2 0 m F N + mg = mv2 r (988 kg) (9.8 m/s) F N = 29. kn (c) From what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop? To do this problem we first want to find the minimum velocity the car must have to stay on the track on the top of the loop, and then use the conservation of energy to find from what height the cart must be released so that it reaches that velocity. To find from what the minimum velocity needed for the cart to stay on the track we must use the newtons second law again, but let the normal force just go to zero. F N + mg = mv2 r So letting F N = 0 and solving for v 2 gives us: v 2 = gr Now we want to use the conservation of energy to find h i, if v 2 f = gr. K + U g = W NC 2 mv2 f 2 mv2 i + mgh f mgh i = 0 mgh i = mgh f + 2 mv2 f = mgh + 2 mgr h i = h f + 2 r = 20 m m h i = 25 m

AP Physics - Chapter 8 Practice Test

AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on

AP1 WEP. Answer: E. The final velocities of the balls are given by v = 2gh.

1. Bowling Ball A is dropped from a point halfway up a cliff. A second identical bowling ball, B, is dropped simultaneously from the top of the cliff. Comparing the bowling balls at the instant they reach

7. Kinetic Energy and Work

Kinetic Energy: 7. Kinetic Energy and Work The kinetic energy of a moving object: k = 1 2 mv 2 Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic

Physics 201 Homework 5

Physics 201 Homework 5 Feb 6, 2013 1. The (non-conservative) force propelling a 1500-kilogram car up a mountain -1.21 10 6 joules road does 4.70 10 6 joules of work on the car. The car starts from rest

Physics 201 Fall 2009 Exam 2 October 27, 2009

Physics 201 Fall 2009 Exam 2 October 27, 2009 Section #: TA: 1. A mass m is traveling at an initial speed v 0 = 25.0 m/s. It is brought to rest in a distance of 62.5 m by a force of 15.0 N. The mass is

1) 0.33 m/s 2. 2) 2 m/s 2. 3) 6 m/s 2. 4) 18 m/s 2 1) 120 J 2) 40 J 3) 30 J 4) 12 J. 1) unchanged. 2) halved. 3) doubled.

Base your answers to questions 1 through 5 on the diagram below which represents a 3.0-kilogram mass being moved at a constant speed by a force of 6.0 Newtons. 4. If the surface were frictionless, the

charge is detonated, causing the smaller glider with mass M, to move off to the right at 5 m/s. What is the

This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring

CHAPTER 6 WORK AND ENERGY

CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From

Chapter 6 Work and Energy

Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system

Curso2012-2013 Física Básica Experimental I Cuestiones Tema IV. Trabajo y energía.

1. A body of mass m slides a distance d along a horizontal surface. How much work is done by gravity? A) mgd B) zero C) mgd D) One cannot tell from the given information. E) None of these is correct. 2.

Conservation of Energy Workshop. Academic Resource Center

Conservation of Energy Workshop Academic Resource Center Presentation Outline Understanding Concepts Kinetic Energy Gravitational Potential Energy Elastic Potential Energy Example Conceptual Situations

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

Assignment Work (Physics) Class :Xi Chapter :04: Motion In PLANE

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Assignment Work (Physics) Class :Xi Chapter :04: Motion In PLANE State law of parallelogram of vector addition and derive expression for resultant of two vectors

C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a

10.1 Quantitative. Answer: A Var: 50+

Chapter 10 Energy and Work 10.1 Quantitative 1) A child does 350 J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 4.8 m above the ground. What is the mass

VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

WORK DONE BY A CONSTANT FORCE

WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of

Physics 2101, First Exam, Fall 2007

Physics 2101, First Exam, Fall 2007 September 4, 2007 Please turn OFF your cell phone and MP3 player! Write down your name and section number in the scantron form. Make sure to mark your answers in the

Ground Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan

PC1221 Fundamentals of Physics I Lectures 9 and 10 he Laws of Motion Dr ay Seng Chuan 1 Ground Rules Switch off your handphone and pager Switch off your laptop computer and keep it No talking while lecture

Work, Kinetic Energy and Potential Energy

Chapter 6 Work, Kinetic Energy and Potential Energy 6.1 The Important Stuff 6.1.1 Kinetic Energy For an object with mass m and speed v, the kinetic energy is defined as K = 1 2 mv2 (6.1) Kinetic energy

B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B

Practice Test 1 1) Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.60 s and reaches the level of the top of the pole after a total elapsed time

KE =? v o. Page 1 of 12

Page 1 of 12 CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be

= Ps cos 0 = (150 N)(7.0 m) = J F N. s cos 180 = µ k

Week 5 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions o these problems, various details have been changed, so that the answers will come out dierently. The method to ind the solution

Mechanics 1. Revision Notes

Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....

Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9

PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always

University Physics 226N/231N Old Dominion University. Newton s Laws and Forces Examples

University Physics 226N/231N Old Dominion University Newton s Laws and Forces Examples Dr. Todd Satogata (ODU/Jefferson Lab) satogata@jlab.org http://www.toddsatogata.net/2012-odu Wednesday, September

Springs. Spring can be used to apply forces. Springs can store energy. These can be done by either compression, stretching, or torsion.

Work-Energy Part 2 Springs Spring can be used to apply forces Springs can store energy These can be done by either compression, stretching, or torsion. Springs Ideal, or linear springs follow a rule called:

AP Physics Newton's Laws Practice Test

AP Physics Newton's Laws Practice Test Answers: A,D,C,D,C,E,D,B,A,B,C,C,A,A 15. (b) both are 2.8 m/s 2 (c) 22.4 N (d) 1 s, 2.8 m/s 16. (a) 12.5 N, 3.54 m/s 2 (b) 5.3 kg 1. Two blocks are pushed along a

Conservative vs. Non-conservative forces Gravitational Potential Energy. Work done by non-conservative forces and changes in mechanical energy

Next topic Conservative vs. Non-conservative forces Gravitational Potential Energy Mechanical Energy Conservation of Mechanical energy Work done by non-conservative forces and changes in mechanical energy

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) The following four forces act on a 4.00 kg object: 1) F 1 = 300 N east F 2 = 700 N north

Ch 6 Forces. Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79

Ch 6 Forces Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79 Friction When is friction present in ordinary life? - car brakes - driving around a turn - walking - rubbing your hands together

PHYS101 The Laws of Motion Spring 2014

The Laws of Motion 1. An object of mass m 1 = 55.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m 2

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

Motion and Forces in Two Dimensions Sec. 7.1 Forces in Two Dimensions 1. A Review of Vector Addition. Forces on an Inclined Plane 3. How to find an Equilibrant Vector 4. Projectile Motion Objectives Determine

Physics 101 Prof. Ekey. Chapter 5 Force and motion (Newton, vectors and causing commotion)

Physics 101 Prof. Ekey Chapter 5 Force and motion (Newton, vectors and causing commotion) Goal of chapter 5 is to establish a connection between force and motion This should feel like chapter 1 Questions

6: Applications of Newton's Laws

6: Applications of Newton's Laws Friction opposes motion due to surfaces sticking together Kinetic Friction: surfaces are moving relative to each other a.k.a. Sliding Friction Static Friction: surfaces

Physics Midterm Review. Multiple-Choice Questions

Physics Midterm Review Multiple-Choice Questions 1. A train moves at a constant velocity of 90 km/h. How far will it move in 0.25 h? A. 10 km B. 22.5 km C. 25 km D. 45 km E. 50 km 2. A bicyclist moves

PHYSICS 111 HOMEWORK#6 SOLUTION. February 22, 2013

PHYSICS 111 HOMEWORK#6 SOLUTION February 22, 2013 0.1 A block of mass m = 3.20 kg is pushed a distance d = 4.60 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

At the skate park on the ramp

At the skate park on the ramp 1 On the ramp When a cart rolls down a ramp, it begins at rest, but starts moving downward upon release covers more distance each second When a cart rolls up a ramp, it rises

PHY231 Section 1, Form B March 22, 2012

1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law

Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION

1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is

Work, Power, Energy Multiple Choice. PSI Physics. Multiple Choice Questions

Work, Power, Energy Multiple Choice PSI Physics Name Multiple Choice Questions 1. A block of mass m is pulled over a distance d by an applied force F which is directed in parallel to the displacement.

f max s = µ s N (5.1)

Chapter 5 Forces and Motion II 5.1 The Important Stuff 5.1.1 Friction Forces Forces which are known collectively as friction forces are all around us in daily life. In elementary physics we discuss the

Explaining Motion:Forces

Explaining Motion:Forces Chapter Overview (Fall 2002) A. Newton s Laws of Motion B. Free Body Diagrams C. Analyzing the Forces and Resulting Motion D. Fundamental Forces E. Macroscopic Forces F. Application

Name: Partners: Period: Coaster Option: 1. In the space below, make a sketch of your roller coaster.

1. In the space below, make a sketch of your roller coaster. 2. On your sketch, label different areas of acceleration. Put a next to an area of negative acceleration, a + next to an area of positive acceleration,

Forces: Equilibrium Examples

Physics 101: Lecture 02 Forces: Equilibrium Examples oday s lecture will cover extbook Sections 2.1-2.7 Phys 101 URL: http://courses.physics.illinois.edu/phys101/ Read the course web page! Physics 101:

Physics 271 FINAL EXAM-SOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath

Physics 271 FINAL EXAM-SOLUTIONS Friday Dec 23, 2005 Prof. Amitabh Lath 1. The exam will last from 8:00 am to 11:00 am. Use a # 2 pencil to make entries on the answer sheet. Enter the following id information

AP Physics C Fall Final Web Review

Name: Class: _ Date: _ AP Physics C Fall Final Web Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. On a position versus time graph, the slope of

Chapter 8: Conservation of Energy

Chapter 8: Conservation of Energy This chapter actually completes the argument established in the previous chapter and outlines the standing concepts of energy and conservative rules of total energy. I

8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential

8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential energy, e.g. a ball in your hand has more potential energy

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was

What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact?

Chapter 4: Forces What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact? Application different forces (field forces, contact

Work-Energy Bar Charts

Name: Work-Energy Bar Charts Read from Lesson 2 of the Work, Energy and Power chapter at The Physics Classroom: http://www.physicsclassroom.com/class/energy/u5l2c.html MOP Connection: Work and Energy:

1) A 2) B 3) C 4) A and B 5) A and C 6) B and C 7) All of the movies A B C. PHYS 11: Chap. 2, Pg 2

1) A 2) B 3) C 4) A and B 5) A and C 6) B and C 7) All of the movies A B C PHYS 11: Chap. 2, Pg 2 1 1) A 2) B 3) C 4) A and B 5) A and C 6) B and C 7) All three A B PHYS 11: Chap. 2, Pg 3 C 1) more than

Work and Direction. Work and Direction. Work and Direction. Work and Direction

Calculate the net gravitational force on the shaded ball. Be sure to include the magnitude and direction. Each ball has a mass of 20,000 kg. (0.79N, 22.5 o N of E) Chapter Six Work = Force X distance W

TEACHER ANSWER KEY November 12, 2003. Phys - Vectors 11-13-2003

Phys - Vectors 11-13-2003 TEACHER ANSWER KEY November 12, 2003 5 1. A 1.5-kilogram lab cart is accelerated uniformly from rest to a speed of 2.0 meters per second in 0.50 second. What is the magnitude

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in

Center of Mass/Momentum

Center of Mass/Momentum 1. 2. An L-shaped piece, represented by the shaded area on the figure, is cut from a metal plate of uniform thickness. The point that corresponds to the center of mass of the L-shaped

Physics General Physics I

Physics 1114 - General Physics I Midterm Solutions 1. Your cat drops from a shelf 1.22 m above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 cm. Ignoring air resistance,

B) 286 m C) 325 m D) 367 m Answer: B

Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of

Work, Energy and Power

Work, Energy and Power In this section of the Transport unit, we will look at the energy changes that take place when a force acts upon an object. Energy can t be created or destroyed, it can only be changed

1 of 7 10/2/2009 1:13 PM

1 of 7 10/2/2009 1:13 PM Chapter 6 Homework Due: 9:00am on Monday, September 28, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap.

This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension PHYS 2: Chap. 19, Pg 2 1 New Topic Phys 1021 Ch 7, p 3 A 2.0 kg wood box slides down a vertical

1. Newton s Laws of Motion and their Applications Tutorial 1

1. Newton s Laws of Motion and their Applications Tutorial 1 1.1 On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 5.00 kg, and on this particular planet its weight is 40.0

Vectors and the Inclined Plane

Vectors and the Inclined Plane Introduction: This experiment is designed to familiarize you with the concept of force as a vector quantity. The inclined plane will be used to demonstrate how one force

Physics-1 Recitation-3

Physics-1 Recitation-3 The Laws of Motion 1) The displacement of a 2 kg particle is given by x = At 3/2. In here, A is 6.0 m/s 3/2. Find the net force acting on the particle. (Note that the force is time

Physics 201 Homework 8

Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

AP Physics B Free Response Solutions

AP Physics B Free Response Solutions. (0 points) A sailboat at rest on a calm lake has its anchor dropped a distance of 4.0 m below the surface of the water. The anchor is suspended by a rope of negligible

SOLUTIONS TO PROBLEM SET 4

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01X Fall Term 2002 SOLUTIONS TO PROBLEM SET 4 1 Young & Friedman 5 26 A box of bananas weighing 40.0 N rests on a horizontal surface.

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,

Solution Derivations for Capa #10

Solution Derivations for Capa #10 1) The flywheel of a steam engine runs with a constant angular speed of 172 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel

Work Energy & Power. September 2000 Number 05. 1. Work If a force acts on a body and causes it to move, then the force is doing work.

PhysicsFactsheet September 2000 Number 05 Work Energy & Power 1. Work If a force acts on a body and causes it to move, then the force is doing work. W = Fs W = work done (J) F = force applied (N) s = distance

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular

5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

5. Forces and Motion-I 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will

103 PHYS - CH7 - Part2

Work due to friction If friction is involved in moving objects, work has to be done against the kinetic frictional force. This work is: W = f d = f d cos180 = f d o f k k k Physics 1 Example f n r θ F

Work. Work = Force x parallel distance (parallel component of displacement) F v

Work Work = orce x parallel distance (parallel component of displacement) W k = d parallel d parallel Units: N m= J = " joules" = ( kg m2/ s2) = average force computed over the distance r r When is not

PHYSICS 218 EXAM 2 Thursday, October 22, 2009

PHYSICS 218 EXAM 2 Thursday, October 22, 2009 NAME: SECTION: 525 526 527 528 Note: 525 Recitation Wed 9:10-10:00 526 Recitation Wed 11:30-12:20 527 Recitation Wed 1:50-2:40 528 Recitation Mon 11:30-12:20

HW#4b Page 1 of 6. I ll use m = 100 kg, for parts b-c: accelerates upwards, downwards at 5 m/s 2 A) Scale reading is the same as person s weight (mg).

HW#4b Page 1 of 6 Problem 1. A 100 kg person stands on a scale. a.) What would be the scale readout? b.) If the person stands on the scale in an elevator accelerating upwards at 5 m/s, what is the scale

F mg (10.1 kg)(9.80 m/s ) m

Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

PHYSICS MIDTERM REVIEW

1. The acceleration due to gravity on the surface of planet X is 19.6 m/s 2. If an object on the surface of this planet weighs 980. newtons, the mass of the object is 50.0 kg 490. N 100. kg 908 N 2. If

Recap. A force is the product of an object s mass and acceleration. Forces are the reason why objects change their velocity. Newton s second law:

Recap A force is the product of an object s mass and acceleration. Forces are the reason why objects change their velocity. Newton s second law: Unit: 1 N = 1 kg m/s 2 Forces are vector quantities, since

PHY121 #8 Midterm I 3.06.2013

PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

P113 University of Rochester NAME S. Manly Fall 2013

Final Exam (December 19, 2013) Please read the problems carefully and answer them in the space provided. Write on the back of the page, if necessary. Show all your work. Partial credit will be given unless

physics 111N forces & Newton s laws of motion

physics 111N forces & Newton s laws of motion forces (examples) a push is a force a pull is a force gravity exerts a force between all massive objects (without contact) (the force of attraction from the

Lecture 9. Friction in a viscous medium Drag Force Quantified

Lecture 9 Goals Describe Friction in Air (Ch. 6) Differentiate between Newton s 1 st, 2 nd and 3 rd Laws Use Newton s 3 rd Law in problem solving Assignment: HW4, (Chap. 6 & 7, due 10/5) 1 st Exam Thurs.,

56 Chapter 5: FORCE AND MOTION I

Chapter 5: FORCE AND MOTION I 1 An example of an inertial reference frame is: A any reference frame that is not accelerating B a frame attached to a particle on which there are no forces C any reference

2. (P2.1 A) a) A car travels 150 km in 3 hours, what is the cars average speed?

Physics: Review for Final Exam 1 st Semester Name Hour P2.1A Calculate the average speed of an object using the change of position and elapsed time 1. (P2.1 A) What is your average speed if you run 140

Practice Test SHM with Answers MPC 1) If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? (There could be more than one

Rotation, Angular Momentum

This test covers rotational motion, rotational kinematics, rotational energy, moments of inertia, torque, cross-products, angular momentum and conservation of angular momentum, with some problems requiring

Announcements. Dry Friction

Announcements Dry Friction Today s Objectives Understand the characteristics of dry friction Draw a FBD including friction Solve problems involving friction Class Activities Applications Characteristics