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


 Samantha Webb
 2 years ago
 Views:
Transcription
1 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 is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem : A 0.kg uniform board is wedged into a corner and held by a spring at a 50.0 angle, as the drawing shows. The spring has a spring constant of 76 N/m and is parallel to the floor. Find the amount by which the spring is stretched from its unstrained length. REASONING AND SOLUTION In order to find the amount the spring stretches we need to calculate the force that acts on the spring. The magnitude of this force is F. Since the board is in equilibrium, the net torque acting on it is zero. Taking the axis of rotation to be at the corner and assuming the board has a length L, the net torque is Στ FL sin 50.0 mg (L/) cos Solving for F gives mg cos 50.0 mg F sin 50.0 tan 50.0 The amount x by which the spring stretches is equal to the magnitude F of the force applied to it divided by the spring constant k (see Equation 0.). Thus, F mg (0. kg)(9.80 m/s ) x k k tan 50.0 (76 N/m) tan m WebAssign Problem : Concept Simulation 0.3 at illustrates the concepts pertinent to this problem. An 0.80kg object is attached to one end of a spring, as in Figure 0.6, and the system is set into simple harmonic motion. The displacement x of the object as a function of time is shown in the drawing. With the aid of these data, determine (a) the amplitude A of the motion, (b) the angular frequency ω (c) the spring constant k, (d) the speed of the object at t.0 s, and (e) the magnitude of the object s acceleration at t.0 s.
2 REASONING AND SOLUTION a. Since the object oscillates between ± m, the amplitude of the motion is m. 0.4, b. From the graph, the period is T 4.0 s. Therefore, according to Equation ω π T π.6 rad/s 4.0 s c. Equation 0. relates the angular frequency to the spring constant: ω k/ m. Solving for k we find k ω m (.6 rad/s) (0.80 kg).0 N/m d. At t.0 s, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is v 0 m/s. e. The acceleration of the object at t.0 s is a maximum, and its magnitude is a max Aω (0.080 m)(.6 rad/s) 0.0 m/s WebAssign Problem 3: A 3.0kg block is between two horizontal springs. Neither spring is strained when the block is at the position labeled x 0 m in the drawing. The block is then displaced a distance of m from the position where x 0 m and is released from rest. (a) What is the speed of the block when it passes back through the x 0 m position? (b) Determine the angular frequency ω of this system. REASONING AND SOLUTION a. When the block passes through the position x 0 m, its velocity is a maximum and can be found from Equation 0.8: v max Aω.
3 We can find the angular frequency ω from the following reasoning: When the mass is given a displacement x, one spring is stretched by an amount x, while the other is compressed by an amount x. The total restoring force on the mass is, therefore, F x k x k x (k + k )x Comparison with Equation 0. shows that the twospring system has an effective spring constant of k eff k + k. Thus, from Equation 0. ω k k + k m m eff Combining this with Equation 0.8 we obtain v max k + k 650 N/m N/m A (0.070 m).3 m/s m 3.0 kg b. The angular frequency of the system is ω k + k 650 N/m N/m m 3.0 kg 9 rad/s WebAssign Problem 4: Concept Simulation 0. at allows you to explore the concepts to which this problem relates. A.00kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.00 m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distances above the point of release, where h 0 m. h PE PE (meters) KE (gravity) (elastic) E REASONING AND SOLUTION If we neglect air resistance, only the conservative forces of the spring and gravity act on the ball. Therefore, the principle of conservation of mechanical energy applies.
4 When the.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount y, where F mg (.00 kg)(9.80 m/s ) y 0.39 m k k 50.0 N/m (0.) This position represents the equilibrium position of the system with the.00kg object suspended from the spring. The object is then pulled down another 0.00 m and released from rest ( v 0 0 m/s). At this point the spring is stretched by an amount of 0.39 m m 0.59 m. This point represents the zero reference level ( h 0 m) for the gravitational potential energy. h 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are: mv 0 m KE (0 m/s) 0 J PE mgh mg(0 m) 0 J gravity elastic ky 0 PE (50.0 N/m)(0.59 m) 8.76 J The total mechanical energy E 0 at the point of release is the sum of the three energies above: E J. h 0.00 m: When the object has risen a distance of h 0.00 m above the release point, the spring is stretched by an amount of 0.59 m 0.00 m 0.39 m. Since the total mechanical energy is conserved, its value at this point is still E 8.76 J. The gravitational and elastic potential energies are: PE mgh (.00 kg)(9.80 m/s )(0.00 m) 3.9 J gravity elastic ky PE (50.0 N/m)(0.39 m) 3.84 J Since KE + PEgravity + PEelastic E, KE E PE PE 8.76 J 3.9 J 3.84 J.00 J gravity elastic h m: When the object has risen a distance of h m above the release point, the spring is stretched by an amount of 0.59 m m 0.9 m. At
5 this point, the total mechanical energy is still E potential energies are: 8.76 J. The gravitational and elastic PE mgh (.00 kg)(9.80 m/s )(0.400 m) 7.84 J gravity The kinetic energy is elastic ky PE (50.0 N/m)(0.9 m) 0.9 J KE E PE PE 8.76 J 7.84 J 0.9 J 0 J gravity elastic The results are summarized in the table below: h KE PEgrav PEelastic E 0 m 0 J 0 J 8.76 J 8.76 J 0.00 m.00 J 3.9 J 3.84 J 8.76 J m 0.00 J 7.84 J 0.9 J 8.76 J WebAssign Problem 5: United States currency is printed using intaglio presses that generate a printing pressure of. A $0 bill is 6. in. by.6 in. Calculate the magnitude of the force that the printing press applies to one side of the bill. REASONING Pressure is the magnitude of the force applied perpendicularly to a surface divided by the area of the surface, according to Equation.3. The force magnitude, therefore, is equal to the pressure times the area. SOLUTION According to Equation.3, we have c h b gb g 4 6 F PA lb / in. 6. in.. 6 in lb WebAssign Problem 6: Two identical containers are open at the top and are connected at the bottom via a tube of negligible volume and a valve that is closed. Both containers are filled initially to the same height of.00 m, one with water, the other with mercury, as the drawing indicates. The valve is then opened. Water and mercury are immiscible. Determine the fluid level in the left container when equilibrium is reestablished.
6 REASONING AND SOLUTION The mercury, being more dense, will flow from the right container into the left container until the pressure is equalized. Then the pressure at the bottom of the left container will be P ρ w gh w + ρ m gh ml and the pressure at the bottom of the right container will be P ρ m gh mr. Equating gives ρ w gh w + ρ m g(h ml h mr ) 0 () Both liquids are incompressible and immiscible so h w.00 m and h ml + h mr.00 m Using these in () and solving for h ml gives, h ml (/)(.00 ρ w /ρ m ) 0.46 m. So the fluid level in the left container is.00 m m.46 m from the bottom. WebAssign Problem 7: Interactive Solution.33 at presents a model for solving this problem. MultipleConcept Example 8 also presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of. The weight of the input piston is negligible. The radii of the input piston and output plunger are and 0.5 m, respectively. What input force F is needed to support the N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is.30 m above that of the input piston? REASONING We label the input piston as and the output plunger as. When the bottom surfaces of the input piston and output plunger are at the same level, F F A / A, applies. However, this equation is not applicable Equation.5, ( ) when the bottom surface of the output plunger is h.50 m above the input piston. In this case we must use Equation.4, P P + ρ gh, to account for the difference in heights. In either case, we will see that the input force is less than the combined weight of the output plunger and car. SOLUTION a. Using A π r for the circular areas of the piston and plunger, the input force required to support the N weight is
7 3 ( m) A π F F ( N ) 93.0 N A π ( 0.5 m) (.5) b. The pressure P at the input piston is related to the pressure P at the bottom of the output plunger by Equation.4, P P + ρgh, where h is the difference in heights. Setting P F / A F / ( π r ), ( P ) F / π r have F F gh r π r + ρ π r ( N ) π π ( π ) 3 ( m) ( 0.5 m), and solving for F, we (.4) 3 3 ( ) ( ) ( ) π ( ) kg/m 9.80 m/s.30 m m 94.9 N WebAssign Problem 8: A paperweight, when weighed in air, has a weight of. When completely immersed in water, however, it has a weight of. Find the volume of the paperweight. REASONING The paperweight weighs less in water than in air, because of the buoyant force F B of the water. The buoyant force points upward, while the weight points downward, leading to an effective weight in water of W In water W F B. There is also a buoyant force when the paperweight is weighed in air, but it is negligibly small. Thus, from the given weights, we can obtain the buoyant force, which is the weight of the displaced water, according to Archimedes principle. From the weight of the displaced water and the density of water, we can obtain the volume of the water, which is also the volume of the completely immersed paperweight. SOLUTION We have W W F or F W W In water B B In water According to Archimedes principle, the buoyant force is the weight of the displaced water, which is mg, where m is the mass of the displaced water. Using Equation., we can write the mass as the density times the volume or m ρv. Thus, for the buoyant force, we have F W W ρ Vg B In water
8 Solving for the volume and using ρ kg/m 3 for the density of water (see Table.), we find V W W ρ g In water 6. 9 N 4. 3 N c.00 0 kg / m hc9. 80 m / s h 4 m 3 WebAssign Problem 9: Concept Simulation. at reviews the concept that plays the central role in this problem. (a) The volume flow rate in an artery supplying the brain is. If the radius of the artery is 5. mm, determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 3. Assume that the volume flow rate is the same as that in part (a). REASONING a. According to Equation.0, the volume flow rate Q is equal to the product of the crosssectional area A of the artery and the speed v of the blood, Q Av. Since Q and A are known, we can determine v. b. Since the volume flow rate Q through the constriction is the same as the volume flow rate Q in the normal part of the artery, Q Q. We can use this relation to find the blood speed in the constricted region. SOLUTION a. Since the artery is assumed to have a circular crosssection, its crosssectional area is A π, where r is the radius. Thus, the speed of the blood is r Q 6 3 Q m / s A π r v 3 ( 5. 0 m) 4. 0 m/s π (.0) b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q Q. Since Q A v, the blood speed is v Q /A Q /A. We are given that the radius of the constricted part of the artery is onethird that of the normal artery, so r r. Thus, the speed of the blood at the constriction is 3 v Q 6 3 Q Q m / s A π r π r π ( ) ( 5. 0 m) 0.38 m/s
9 WebAssign Problem 0: See MultipleConcept Example 5 to review the concepts that are pertinent to this problem. The blood speed in a normal segment of a horizontal artery is 0. m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to onefourth the normal crosssectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? REASONING We assume that region contains the constriction and region is the normal region. The difference in blood pressures between the two points in the horizontal artery is given by Bernoulli s equation (Equation.) as P P ρ v ρ v, where v and v are the speeds at the two points. Since the volume flow rate is the same at the two points, the speed at is related to the speed at by Equation.9, the equation of continuity: A v A v, where A and A are the crosssectional areas of the artery. By combining these two relations, we will be able to determine the pressure difference. SOLUTION Solving the equation of continuity for the blood speed in region gives v v A /A. Substituting this result into Bernoulli s equation yields Since A 4 v A ρ ρ ρ ρ A P P v v v A, the pressure difference is v A ρ ρ ρ A 4 3 ( ) ( ) ( ) ( ) P P v v kg/m 0. m/s 5 96 Pa We have taken the density ρ of blood from Table.. WebAssign Problem : Poiseuilles' law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about 00. Re R / Here,, and are, respectively, the average speed, density, and viscosity of the fluid, and R is the radius of the pipe. Calculate the highest average speed that blood ( 060 kg/m 3, Pa s) could have and still remain in laminar flow when it flows through the aorta (R m).
10 REASONING AND SOLUTION The Reynold's number, Re, can be written as Re v ρ R / η. To find the average speed v, v 3 ( 000) ( Pa s) 3 3 ( ) ( ) (Re) η ρ R 060 kg/m m 0.5 m/s
11 Practice conceptual problems: Note: Chapter 0 problems were included in last week s solutions. Chapter : 3. A person could not balance her entire weight on the pointed end of a single nail, because it would penetrate her skin. However, she can lie safely on a bed of nails. A bed of nails consists of many nails driven through a sheet of wood so that the pointed ends form a flat array. Why is the bed of nails trick safe? REASONING AND SOLUTION A person could not balance her entire weight on the pointed end of a single nail, because it would penetrate her skin. According to Equation.3, the pressure exerted by the nail is P F / A where F represents the weight of the person, and A is the area of the tip of the nail. Since the tip of the nail has a very small radius, its area is very small; therefore, the pressure that the nail exerts on the person is large. The reason she can safely lie on a "bed of nails" is that the effective area of the nails is very large if the nails are closely spaced. Thus, the weight of the person F is distributed over all the nails so that the pressure exerted by any one nail is small. 8. Could you use a straw to sip a drink on the moon? Explain. REASONING AND SOLUTION When you drink through a straw, you draw the air out of the straw, and the external air pressure leads to the unbalanced force that pushes the liquid up into the straw. This action requires the presence of an atmosphere. The moon has no atmosphere, so you could not use a straw to sip a drink on the moon. 3. On a distant planet the acceleration due to gravity is less than it is on earth. Would you float more easily in water on this planet than on earth? Account for your answer. REASONING AND SOLUTION According to Archimedes' principle, any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces. Therefore, the magnitude of the buoyant force exerted on an object immersed in water is given by F ρ Vg B water, where ρ water is the density of water, V is the volume displaced by the immersed object, and g is the magnitude of the acceleration due to gravity. If the acceleration due to gravity on a distant planet is less than it is on earth, then, other factors remaining the same, the buoyant force will be less on the planet than it is on earth. However, the weight of the object will also be less than the weight of the object on earth.
12 When an object floats in water, the upward buoyant force exerted by the water must be equal in magnitude and opposite in direction to the weight of the object, as shown at the right. Hence, F m g B object. It follows that ρ water Vg m object g. Notice that the acceleration due to gravity, g, appears on both sides of this equation. Algebraically canceling the g's we have ρ water V m object. Therefore, the object will float so that it displaces a volume of water V, where V m / ρ object water. This result is independent of g. It is the same on earth as it is on the distant planet. Therefore, it would be no more difficult to float in water on this planet than it would be on earth. B u o y a n t f o r c e o n o b j e c t W e i g h t o f o b j e c t 8. In steady flow, the velocity of a fluid particle at any point is constant in time. On the other hand, a fluid accelerates when it moves into a region of smaller crosssectional area. (a) Explain what causes the acceleration. (b) Explain why the condition of steady flow does not rule out such an acceleration. REASONING AND SOLUTION In steady flow, the velocity v of a fluid particle at any point is constant in time. On the other hand, a fluid accelerates when it moves into a region of smaller crosssectional area, as shown in the figure below. X v X Y v Y a. A fluid particle at X with speed v X must be accelerated to the right in order to acquire the greater speed v Y at Y. From Newton's second law, this acceleration can arise only from a net force that acts in the direction XY. If there are no other external forces acting on the fluid, this force must arise from the change in pressure within the fluid. The pressure at X must be greater than the pressure at Y. b. The definition of steady flow makes no reference as to how the velocity of a fluid particle varies from point to point as the fluid flows. It simply states that the velocity of a fluid particle at any particular point is constant in time. Therefore, the condition of steady flow does not rule out the acceleration discussed in part (a). 3. Which way would you have to spin a baseball so that it curves upward on its way to the plate? In describing the spin, state how you are viewing the ball. Justify your answer. REASONING AND SOLUTION The figure below shows a baseball, as viewed from the side, moving to the right with no spin. Since the air flows with the same speed above and below the ball, the pressure is reduced by the same amount above
13 and below the ball. There is no net force to cause the ball to curve in any particular direction (except for gravity which results in the usual parabolic trajectory). up v ball right Without spin If the ball is given a spin that is counterclockwise when viewed from the side, as shown below, the air close to the surface of the ball is dragged with the ball. In accord with Bernoulli's equation, the air on the top half of the ball is "speeded up" (pressure reduced by a greater amount), while that on the lower half of the ball is also speeded up, but less so (pressure reduced by a smaller amount). Thus, the pressure on the top half of the ball is lower than on the bottom half. F a s t e r a i r, l o w e r p r e s s u r e D e f l e c t i o n f o r c e u p v b a l l r i g h t S l o w e r a i r, h i g h p r e s s u r e Because of the pressure difference, a deflection force is generated that is directed from the higher pressure side of the ball to the lower pressure side of the ball. Therefore, the ball curves upward on its way to the plate.
IMPORTANT NOTE ABOUT WEBASSIGN:
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
More informationPhysics 41 HW Set 1 Chapter 15
Physics 4 HW Set Chapter 5 Serway 8 th OC:, 4, 7 CQ: 4, 8 P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59, 67, 74 OC CQ P: 4, 5, 8, 8, 0, 9,, 4, 9, 4, 5, 5 Discussion Problems:, 57, 59,
More informationChapter 13, example problems: x (cm) 10.0
Chapter 13, example problems: (13.04) Reading Fig. 1330 (reproduced on the right): (a) Frequency f = 1/ T = 1/ (16s) = 0.0625 Hz. (since the figure shows that T/2 is 8 s.) (b) The amplitude is 10 cm.
More informationA1. An object of mass m is projected vertically from the surface of a planet of radius R p and mass M p with an initial speed v i.
OBAFMI AWOLOWO UNIVRSITY, ILIF, IF, NIGRIA. FACULTY OF SCINC DPARTMNT OF PHYSICS B.Sc. (Physics) Degree xamination PHY GNRAL PHYSICS I TUTORIAL QUSTIONS IN GRAVITATION, FLUIDS AND OSCILLATIONS SCTION
More informationChapter 8 Fluid Flow
Chapter 8 Fluid Flow GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational
More informationLab 5: Conservation of Energy
Lab 5: Conservation of Energy Equipment SWS, 1meter stick, 2meter stick, heavy duty bench clamp, 90cm rod, 40cm rod, 2 double clamps, brass spring, 100g mass, 500g mass with 5cm cardboard square
More informationPractice Test SHM with Answers
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
More informationPHY231 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
More informationCHAPTER 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
More informationPHY121 #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
More informationChapter 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
More informationPhysics 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
More informationPHY231 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
More information3600 s 1 h. 24 h 1 day. 1 day
Week 7 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
More information9. 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
More informationSolution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:
Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is
More information8. 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
More informationCh 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
More informationSample Questions for the AP Physics 1 Exam
Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiplechoice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each
More informationChapter 28 Fluid Dynamics
Chapter 28 Fluid Dynamics 28.1 Ideal Fluids... 1 28.2 Velocity Vector Field... 1 28.3 Mass Continuity Equation... 3 28.4 Bernoulli s Principle... 4 28.5 Worked Examples: Bernoulli s Equation... 7 Example
More informationCH205: Fluid Dynamics
CH05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Solutions of Mid Semester Examination Data Given: Density of water, ρ = 1000 kg/m 3, gravitational acceleration, g
More informationC 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
More informationoil liquid water water liquid Answer, Key Homework 2 David McIntyre 1
Answer, Key Homework 2 David McIntyre 1 This printout should have 14 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making
More informationWork, 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.
More information2.1 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
More informationLecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is
Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.49.6, 10.110.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of
More informationPHYS 1014M, Fall 2005 Exam #3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
PHYS 1014M, Fall 2005 Exam #3 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A bicycle wheel rotates uniformly through 2.0 revolutions in
More informationCurso20122013 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.
More informationHomework 4. problems: 5.61, 5.67, 6.63, 13.21
Homework 4 problems: 5.6, 5.67, 6.6,. Problem 5.6 An object of mass M is held in place by an applied force F. and a pulley system as shown in the figure. he pulleys are massless and frictionless. Find
More informationPhysics 1120: Simple Harmonic Motion Solutions
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Physics 1120: Simple Harmonic Motion Solutions 1. A 1.75 kg particle moves as function of time as follows: x = 4cos(1.33t+π/5) where distance is measured
More informationEDUH 1017  SPORTS MECHANICS
4277(a) Semester 2, 2011 Page 1 of 9 THE UNIVERSITY OF SYDNEY EDUH 1017  SPORTS MECHANICS NOVEMBER 2011 Time allowed: TWO Hours Total marks: 90 MARKS INSTRUCTIONS All questions are to be answered. Use
More informationChapter 13  Solutions
= Chapter 13  Solutions Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron. Part A On a parttime job you are asked to bring a cylindrical iron rod
More informationAP Physics C. Oscillations/SHM Review Packet
AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete
More informationAP1 Oscillations. 1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false?
1. Which of the following statements about a springblock oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The
More informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 Nm is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kgm 2. What is the
More informationv 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
More informationAP 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
More information7. 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
More informationSo if ω 0 increases 3fold, the stopping angle increases 3 2 = 9fold.
Name: MULTIPLE CHOICE: Questions 111 are 5 points each. 1. A safety device brings the blade of a power mower from an angular speed of ω 1 to rest in 1.00 revolution. At the same constant angular acceleration,
More informationForces. Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy
Forces Definition Friction Falling Objects Projectiles Newton s Laws of Motion Momentum Universal Forces Fluid Pressure Hydraulics Buoyancy Definition of Force Force = a push or pull that causes a change
More informationPhysics 112 Homework 5 (solutions) (2004 Fall) Solutions to Homework Questions 5
Solutions to Homework Questions 5 Chapt19, Problem2: (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat
More informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More informationSimple Harmonic Motion
Simple Harmonic Motion 1 Object To determine the period of motion of objects that are executing simple harmonic motion and to check the theoretical prediction of such periods. 2 Apparatus Assorted weights
More informationPhysics 1A Lecture 10C
Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. Oprah Winfrey Static Equilibrium
More information= 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
More informationPhysics 1114: Unit 6 Homework: Answers
Physics 1114: Unit 6 Homework: Answers Problem set 1 1. A rod 4.2 m long and 0.50 cm 2 in crosssectional area is stretched 0.20 cm under a tension of 12,000 N. a) The stress is the Force (1.2 10 4 N)
More information226 Chapter 15: OSCILLATIONS
Chapter 15: OSCILLATIONS 1. In simple harmonic motion, the restoring force must be proportional to the: A. amplitude B. frequency C. velocity D. displacement E. displacement squared 2. An oscillatory motion
More information1 of 10 11/23/2009 6:37 PM
hapter 14 Homework Due: 9:00am on Thursday November 19 2009 Note: To understand how points are awarded read your instructor's Grading Policy. [Return to Standard Assignment View] Good Vibes: Introduction
More information1) 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 3.0kilogram mass being moved at a constant speed by a force of 6.0 Newtons. 4. If the surface were frictionless, the
More informationSpring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations
Spring Simple Harmonic Oscillator Simple Harmonic Oscillations and Resonance We have an object attached to a spring. The object is on a horizontal frictionless surface. We move the object so the spring
More informationB) 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
More informationObjective: Work Done by a Variable Force Work Done by a Spring. Homework: Assignment (125) Do PROBS # (64, 65) Ch. 6, + Do AP 1986 # 2 (handout)
Double Date: Objective: Work Done by a Variable Force Work Done by a Spring Homework: Assignment (125) Do PROBS # (64, 65) Ch. 6, + Do AP 1986 # 2 (handout) AP Physics B Mr. Mirro Work Done by a Variable
More informationChapter 9: The Behavior of Fluids
Chapter 9: The Behavior of Fluids 1. Archimedes Principle states that A. the pressure in a fluid is directly related to the depth below the surface of the fluid. B. an object immersed in a fluid is buoyed
More informationPHYS 211 FINAL FALL 2004 Form A
1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each
More informationNEWTON S LAWS OF MOTION
NEWTON S LAWS OF MOTION Background: Aristotle believed that the natural state of motion for objects on the earth was one of rest. In other words, objects needed a force to be kept in motion. Galileo studied
More informationWORK 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 Newtonmeter (Nm) = Joule, J If you exert a force of
More informationChapter 3.8 & 6 Solutions
Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation 3.26. To find speed we need to know the distance traveled
More informationWeight 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
More informationPhysics 9 Fall 2009 Homework 2  Solutions
Physics 9 Fall 009 Homework  s 1. Chapter 7  Exercise 5. An electric dipole is formed from ±1.0 nc charges spread.0 mm apart. The dipole is at the origin, oriented along the y axis. What is the electric
More informationcharge 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, 2D collisions, and centerofmass, with some problems requiring
More informationChapter 13. Gravitation
Chapter 13 Gravitation 13.2 Newton s Law of Gravitation In vector notation: Here m 1 and m 2 are the masses of the particles, r is the distance between them, and G is the gravitational constant. G = 6.67
More informationChapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued Clicker Question 4.3 A mass at rest on a ramp. How does the friction between the mass and the table know how much force will EXACTLY balance the gravity
More informationNo Brain Too Small PHYSICS. 2 kg
MECHANICS: ANGULAR MECHANICS QUESTIONS ROTATIONAL MOTION (2014;1) Universal gravitational constant = 6.67 10 11 N m 2 kg 2 (a) The radius of the Sun is 6.96 10 8 m. The equator of the Sun rotates at a
More informationHOOKE S LAW AND SIMPLE HARMONIC MOTION
HOOKE S LAW AND SIMPLE HARMONIC MOTION Alexander Sapozhnikov, Brooklyn College CUNY, New York, alexs@brooklyn.cuny.edu Objectives Study Hooke s Law and measure the spring constant. Study Simple Harmonic
More information1. 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
More informationXI / PHYSICS FLUIDS IN MOTION 11/PA
Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A
More informationPhysics 113 Exam #4 Angular momentum, static equilibrium, universal gravitation, fluid mechanics, oscillatory motion (first part)
Physics 113 Exam #4 Angular momentum, static equilibrium, universal gravitation, fluid mechanics, oscillatory motion (first part) Answer all questions on this examination. You must show all equations,
More informationAnnouncements. 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
More informationChapter 24 Physical Pendulum
Chapter 4 Physical Pendulum 4.1 Introduction... 1 4.1.1 Simple Pendulum: Torque Approach... 1 4. Physical Pendulum... 4.3 Worked Examples... 4 Example 4.1 Oscillating Rod... 4 Example 4.3 Torsional Oscillator...
More informationThese slides contain some notes, thoughts about what to study, and some practice problems. The answers to the problems are given in the last slide.
Fluid Mechanics FE Review Carrie (CJ) McClelland, P.E. cmcclell@mines.edu Fluid Mechanics FE Review These slides contain some notes, thoughts about what to study, and some practice problems. The answers
More informationChapter 8: Potential Energy and Conservation of Energy. Work and kinetic energy are energies of motion.
Chapter 8: Potential Energy and Conservation of Energy Work and kinetic energy are energies of motion. Consider a vertical spring oscillating with mass m attached to one end. At the extreme ends of travel
More informationSolution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
More informationAt 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
More informationPhysics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives
Physics 9e/Cutnell correlated to the College Board AP Physics 1 Course Objectives Big Idea 1: Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring
More informationMonday 20 May 2013 Afternoon
Monday 20 May 2013 Afternoon AS GCE PHYSICS A G481/01 Mechanics *G411700613* Candidates answer on the Question Paper. OCR supplied materials: Data, Formulae and Relationships Booklet (sent with general
More informationSimple Harmonic Motion(SHM) Period and Frequency. Period and Frequency. Cosines and Sines
Simple Harmonic Motion(SHM) Vibration (oscillation) Equilibrium position position of the natural length of a spring Amplitude maximum displacement Period and Frequency Period (T) Time for one complete
More informationWORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS
WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS 1. Stored energy or energy due to position is known as Potential energy. 2. The formula for calculating potential energy is mgh. 3. The three factors that
More informationMidterm 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
More informationChapter 11 Equilibrium
11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of
More informationDensity (r) Chapter 10 Fluids. Pressure 1/13/2015
1/13/015 Density (r) Chapter 10 Fluids r = mass/volume Rho ( r) Greek letter for density Units  kg/m 3 Specific Gravity = Density of substance Density of water (4 o C) Unitless ratio Ex: Lead has a sp.
More informationPhysics 117.3 Tutorial #1 January 14 to 25, 2013
Physics 117.3 Tutorial #1 January 14 to 25, 2013 Rm 130 Physics 8.79. The location of a person s centre of gravity can be determined using the arrangement shown in the figure. A light plank rests on two
More informationChapter 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
More informationChapter 4: Newton s Laws: Explaining Motion
Chapter 4: Newton s Laws: Explaining Motion 1. All except one of the following require the application of a net force. Which one is the exception? A. to change an object from a state of rest to a state
More informationCh 7 Kinetic Energy and Work. Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43
Ch 7 Kinetic Energy and Work Question: 7 Problems: 3, 7, 11, 17, 23, 27, 35, 37, 41, 43 Technical definition of energy a scalar quantity that is associated with that state of one or more objects The state
More informationTennessee State University
Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an Fgrade. Other instructions will be given in the Hall. MULTIPLE CHOICE.
More informationPHYSICS 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
More informationSerway_ISM_V1 1 Chapter 4
Serway_ISM_V1 1 Chapter 4 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Newton s second law gives the net force acting on the crate as This gives the kinetic friction force as, so choice (a) is correct. 2. As
More informationKinetic Energy (A) stays the same stays the same (B) increases increases (C) stays the same increases (D) increases stays the same.
1. A cart full of water travels horizontally on a frictionless track with initial velocity v. As shown in the diagram, in the back wall of the cart there is a small opening near the bottom of the wall
More informationPrelab Exercises: Hooke's Law and the Behavior of Springs
59 Prelab Exercises: Hooke's Law and the Behavior of Springs Study the description of the experiment that follows and answer the following questions.. (3 marks) Explain why a mass suspended vertically
More informationPhysics: 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 buckleup? A) the first law
More informationFORCES AND MOTION UNIT TEST. Multiple Choice: Draw a Circle Completely around the ONE BEST answer.
FORCES AND MOTION UNIT TEST Multiple Choice: Draw a Circle Completely around the ONE BEST answer. 1. A force acting on an object does no work if a. a machine is used to move the object. b. the force is
More informationPhysics 200A FINALS Shankar 180mins December 13, 2005 Formulas and figures at the end. Do problems in 4 books as indicated
1 Physics 200A FINALS Shankar 180mins December 13, 2005 Formulas and figures at the end. Do problems in 4 books as indicated I. Book I A camper is trying to boil water. The 55 g aluminum pan has specific
More informationExemplar Problems Physics
Chapter Eight GRAVITATION MCQ I 8.1 The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration
More informationLecture L222D Rigid Body Dynamics: Work and Energy
J. Peraire, S. Widnall 6.07 Dynamics Fall 008 Version.0 Lecture L  D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L3 for
More informationNewton s Laws of Motion
Physics Newton s Laws of Motion Newton s Laws of Motion 4.1 Objectives Explain Newton s first law of motion. Explain Newton s second law of motion. Explain Newton s third law of motion. Solve problems
More informationApplied Fluid Mechanics
Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and
More informationHigher Technological Institute Civil Engineering Department. Lectures of. Fluid Mechanics. Dr. Amir M. Mobasher
Higher Technological Institute Civil Engineering Department Lectures of Fluid Mechanics Dr. Amir M. Mobasher 1/14/2013 Fluid Mechanics Dr. Amir Mobasher Department of Civil Engineering Faculty of Engineering
More informationChapter 11. h = 5m. = mgh + 1 2 mv 2 + 1 2 Iω 2. E f. = E i. v = 4 3 g(h h) = 4 3 9.8m / s2 (8m 5m) = 6.26m / s. ω = v r = 6.
Chapter 11 11.7 A solid cylinder of radius 10cm and mass 1kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30 degrees (a) What is the angular speed
More informationGrade 8 Science Chapter 9 Notes
Grade 8 Science Chapter 9 Notes Force Force  Anything that causes a change in the motion of an object.  usually a push or a pull.  the unit for force is the Newton (N). Balanced Forces  forces that
More information