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9 Week 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 1: A 10.1-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 176 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 = FL sin 50.0 - mg (L/2) cos 50.0 = 0 Solving for F gives F= mg cos 50.0 mg = 2 sin 50.0 2 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 10.1). Thus, x= F mg (10.1 kg)(9.80 m/s 2 ) = = = 0.236 m k 2k tan 50.0 2(176 N/m) tan 50.0 WebAssign Problem 2: Concept Simulation 10.3 at www.wiley.com/college/cutnell illustrates the concepts pertinent to this problem. An 0.80-kg object is attached to one end of a spring, as in Figure 10.6, 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 (c) the spring constant k, (d) the speed of the object at t = 1.0 s, and (e) the magnitude of the object's acceleration at t = 1.0 s. REASONING AND SOLUTION a. Since the object oscillates between 0.080 m , the amplitude of the motion is 0.080 m . b. From the graph, the period is T = 4.0 s . Therefore, according to Equation 10.4, 2 2 = = 1.6 rad/s T 4.0 s = = c. Equation 10.11 relates the angular frequency to the spring constant: k / m . Solving for k we find k = 2 m = (1.6 rad/s) 2 (0.80 kg) = 2.0 N/m d. At t = 1.0 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 = 1.0 s is a maximum, and its magnitude is a max = A 2 = (0.080 m)(1.6 rad/s) 2 = 0.20 m/s 2 WebAssign Problem 3: A 3.0-kg 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 0.070 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 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 10.8: vmax = A. We can find the angular frequency 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, Fx = -k1x - k2x = -(k1 + k2)x Comparison with Equation 10.2 shows that the two-spring system has an effective spring constant of keff = k1 + k2. Thus, from Equation 10.11 = keff m = k1 + k2 m Combining this with Equation 10.8 we obtain vmax = A k1 + k2 650 N/m + 450 N/m = (0.070 m) = 1.3 m/s m 3.0 kg b. The angular frequency of the system is = k1 + k2 = m 650 N/m + 450 N/m = 19 rad/s 3.0 kg WebAssign Problem 4: Concept Simulation 10.1 at www.wiley.com/college/cutnell allows you to explore the concepts to which this problem relates. A 2.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.200 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 0 0.200 0.400 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. When the 2.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount y, where y= F mg (2.00 kg)(9.80 m/s 2 ) = = = 0.392 m k k 50.0 N/m (10.1) This position represents the equilibrium position of the system with the 2.00-kg object suspended from the spring. The object is then pulled down another 0.200 m and released from rest ( v0 = 0 m/s). At this point the spring is stretched by an amount of 0.392 m + 0.200 m = 0.592 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: KE = 1 mv 2 0 2 = 1 m(0 m/s) 2 2 = 0J PE gravity = mgh = mg (0 m) = 0 J PE elastic = 1 ky 2 2 0 = 1 (50.0 2 N/m)(0.592 m) 2 = 8.76 J The total mechanical energy E0 at the point of release is the sum of the three energies above: E0 = 8.76 J . h = 0.200 m: When the object has risen a distance of h = 0.200 m above the release point, the spring is stretched by an amount of 0.592 m 0.200 m = 0.392 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 gravity = mgh = (2.00 kg)(9.80 m/s 2 )(0.200 m) = 3.92 J PE elastic = 1 ky 2 2 = 1 (50.0 2 N/m)(0.392 m) 2 = 3.84 J Since KE + PE gravity + PE elastic = E , KE = E PE gravity PE elastic = 8.76 J 3.92 J 3.84 J = 1.00 J h = 0.400 m: When the object has risen a distance of h = 0.400 m above the release point, the spring is stretched by an amount of 0.592 m 0.400 m = 0.192 m . At this point, the total mechanical energy is still E = 8.76 J . The gravitational and elastic potential energies are: PE gravity = mgh = (2.00 kg)(9.80 m/s 2 )(0.400 m) = 7.84 J PE elastic = The kinetic energy is KE = E PE gravity PE elastic = 8.76 J 7.84 J 0.92 J = 0 J The results are summarized in the table below: h 0m 0.200 m 0.400 m KE 0J 1.00 J 0.00 J PEgrav 0J 3.92 J 7.84 J PEelastic 8.76 J 3.84 J 0.92 J E 8.76 J 8.76 J 8.76 J 1 ky 2 2 = 1 (50.0 2 N/m)(0.192 m) 2 = 0.92 J WebAssign Problem 5: United States currency is printed using intaglio presses that generate a printing pressure of .2 A $20 bill is 6.1 in. by 2.6 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 11.3. The force magnitude, therefore, is equal to the pressure times the area. SOLUTION According to Equation 11.3, we have F = PA = 8.0 10 4 lb / in. 2 c h b 6.1 in.gb 2.6 in.g = 1.3 10 6 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 1.00 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. 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 = wghw + mghmL and the pressure at the bottom of the right container will be P = mghmR. Equating gives wghw + mg(hmL - hmR) = 0 Both liquids are incompressible and immiscible so hw = 1.00 m and hmL + hmR = 1.00 m Using these in (1) and solving for hmL gives, hmL = (1/2)(1.00 - w/m) = 0.46 m. (1) So the fluid level in the left container is 1.00 m + 0.46 m = 1.46 m from the bottom. WebAssign Problem 7: Interactive Solution 11.33 at www.wiley.com/college/cutnell presents a model for solving this problem. Multiple-Concept 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 0.125 m, respectively. What input force F is needed to support the 24 500-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 1.30 m above that of the input piston? REASONING We label the input piston as "2" and the output plunger as "1." When the bottom surfaces of the input piston and output plunger are at the same level, Equation 11.5, F2 = F1 ( A2 / A1 ) , applies. However, this equation is not applicable when the bottom surface of the output plunger is h = 1.50 m above the input piston. In this case we must use Equation 11.4, P2 = P + gh , to account for the difference 1 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 2 a. Using A = r for the circular areas of the piston and plunger, the input force required to support the 24 500-N weight is -3 A2 7.70 10 m F2 = F1 2 A = ( 24 500 N ) 1 ( 0.125 m ) ( ) 2 = 93.0 N (11.5) b. The pressure P2 at the input piston is related to the pressure P1 at the bottom of the output plunger by Equation 11.4, P2 = P1 + gh, where h is the difference in 2 2 heights. Setting P2 = F2 / A2 = F2 / r2 , P = F1 / r1 , and solving for F2, we 1 ( ) ( ) have r22 F2 = F1 2 + gh r22 r 1 ( ) ) 2 (11.4) -3 7.70 10 m = ( 24 500 N ) 2 ( 0.125 m ) ( + 8.30 102 kg/m3 9.80 m/s 2 ( 1.30 m ) 7.70 10- 3 m ( )( ) ( ) 2 = 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 the volume of the paperweight. . . Find REASONING The paperweight weighs less in water than in air, because of the buoyant force FB of the water. The buoyant force points upward, while the weight points downward, leading to an effective weight in water of WIn water = W FB. 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 weight the 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 WIn water = W - FB or FB = W - WIn 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 11.1, we can write the mass as the density times the volume or m = V. Thus, for the buoyant force, we have FB = W - WIn water = Vg Solving for the volume and using = 1.00 10 kg/m for the density of water (see Table 11.1), we find V= W - WIn water = 3 3 g c 6 .9 N - 4 .3 N 1.00 10 3 kg / m 3 9 .80 m / s 2 hc h = 2 .7 10 - 4 m 3 WebAssign Problem 9: Concept Simulation 11.1 at www.wiley.com/college/cutnell 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 5.2 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 11.10, the volume flow rate Q is equal to the product of the cross-sectional 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 Q2 through the constriction is the same as the volume flow rate Q1 in the normal part of the artery, Q2 = Q1. 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 cross-section, its cross-sectional area is 2 , where r is the radius. Thus, the speed of the blood is A1 = r1 1 v1 = Q1 Q1 3.6 10- 6 m3 / s = = = 4.2 10- 2 m/s 2 2 A1 r1 5.2 10- 3 m ( ) (11.10) b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q2 = Q1. Since Q2 = A2v2 , the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the radius of the constricted part of the artery is one-third that of the normal artery, so r2 = 1 r1. Thus, the speed of the blood at the constriction is 3 v2 = Q1 Q Q1 = 12 = A2 r2 1 r 3 1 ( ) 2 = 3.6 10- 6 m3 / s 1 5.2 10- 3 m 3 ( ) 2 = 0.38 m/s WebAssign Problem 10: See Multiple-Concept Example 15 to review the concepts that are pertinent to this problem. The blood speed in a normal segment of a horizontal artery is 0.11 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? REASONING We assume that region 1 contains the constriction and region 2 is the normal region. The difference in blood pressures between the two points in the horizontal artery is given by Bernoulli's equation (Equation 11.12) as 2 2 P2 - P = 1 v1 - 1 v2 , where v1 and v2 are the speeds at the two points. Since 1 2 2 the volume flow rate is the same at the two points, the speed at 1 is related to the speed at 2 by Equation 11.9, the equation of continuity: A1v1 = A2v2, where A1 and A2 are the cross-sectional 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 1 gives v1 = v2A2/A1. Substituting this result into Bernoulli's equation yields P2 - P = 1 Since A1 = 1 4 1 2 2 v1 - 1 2 2 v2 = 1 2 v A 2 2 A 1 2 - 1 2 2 v2 A2 , the pressure difference is v A P2 - P = 1 12 2 - 1 2 4 A2 = 1 2 2 1 2 2 v2 = 1 2 2 v2 ( 16 - 1) ( 1060 kg/m3 ) ( 0.11 m/s ) 2 ( 15) = 96 Pa We have taken the density of blood from Table 11.1. WebAssign Problem 11: 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 2100. Re = 2 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 ( = 1060 kg/m3, = 4.0 10-3 Pa s) could have and still remain in laminar flow when it flows through the aorta (R = 8.0 10-3 m). REASONING AND SOLUTION The Reynold's number, Re, can be written as Re = 2v R / . To find the average speed v , ( 2000 ) 4.0 10- 3 Pa s (Re) v= = = 0.5 m/s 2 R 2 1060 kg/m3 8.0 10- 3 m ( ( )( ) ) Practice conceptual problems: Note: Chapter 10 problems were included in last week's solutions. Chapter 11: 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 11.3, 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. 13. 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 FB = water Vg , where 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. 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, FB = mobject g . It follows that water Vg = mobject g . Notice that the acceleration due to gravity, g, appears on both sides of this equation. Algebraically canceling the g's we have water V = mobject . Therefore, the object will float so that it displaces a volume of water V, where V = mobject / 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 fo rc e o n o b je c t W e i g h t o f o b je c t 18. 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 cross-sectional 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 cross-sectional area, as shown in the figure below. X vX Y vY a. A fluid particle at X with speed vX must be accelerated to the right in order to acquire the greater speed vY 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). 23. 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 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 ste r a ir, l o w e r p re ssu re D e f l e c t io n f o r c e up v b a ll r ig h t S l o w e r a ir, h i g h p re ssu re 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. ... View Full Document