Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 13, Problem 66PQ
To give a pet hamster exercise, some people put the hamster in a ventilated ball and allow it roam around the house (Fig. P13.66). When a hamster is in such a ball, it can cross a typical room in a few minutes. Estimate the total kinetic energy in the ball-hamster system.
FIGURE P13.66 Problems 66 and 67
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Physics for Scientists and Engineers: Foundations and Connections
Ch. 13.1 - CASE STUDY When Is Energy Conserved? Under what...Ch. 13.6 - Figure 13.24 shows a particle with momentum p....Ch. 13.7 - Prob. 13.3CECh. 13.7 - Prob. 13.4CECh. 13.7 - Prob. 13.5CECh. 13 - Prob. 1PQCh. 13 - Prob. 2PQCh. 13 - A Frisbee flies across a field. Determine if the...Ch. 13 - Prob. 4PQCh. 13 - Prob. 5PQ
Ch. 13 - Rotational Inertia Problems 5 and 6 are paired. 5....Ch. 13 - A 12.0-kg solid sphere of radius 1.50 m is being...Ch. 13 - A figure skater clasps her hands above her head as...Ch. 13 - A solid sphere of mass M and radius Ris rotating...Ch. 13 - Suppose a disk having massMtot and radius R is...Ch. 13 - Problems 11 and 12 are paired. A thin disk of...Ch. 13 - Given the disk and density in Problem 11, derive...Ch. 13 - A large stone disk is viewed from above and is...Ch. 13 - Prob. 14PQCh. 13 - A uniform disk of mass M = 3.00 kg and radius r =...Ch. 13 - Prob. 16PQCh. 13 - Prob. 17PQCh. 13 - The system shown in Figure P13.18 consisting of...Ch. 13 - A 10.0-kg disk of radius 2.0 m rotates from rest...Ch. 13 - Prob. 20PQCh. 13 - Prob. 21PQCh. 13 - In Problem 21, what fraction of the kinetic energy...Ch. 13 - Prob. 23PQCh. 13 - Prob. 24PQCh. 13 - Prob. 25PQCh. 13 - A student amuses herself byspinning her pen around...Ch. 13 - The motion of spinning a hula hoop around one's...Ch. 13 - Prob. 28PQCh. 13 - Prob. 29PQCh. 13 - Prob. 30PQCh. 13 - Sophia is playing with a set of wooden toys,...Ch. 13 - Prob. 32PQCh. 13 - A spring with spring constant 25 N/m is compressed...Ch. 13 - Prob. 34PQCh. 13 - Prob. 35PQCh. 13 - Prob. 36PQCh. 13 - Prob. 37PQCh. 13 - Prob. 38PQCh. 13 - A parent exerts a torque on a merry-go-round at a...Ch. 13 - Prob. 40PQCh. 13 - Today, waterwheels are not often used to grind...Ch. 13 - Prob. 42PQCh. 13 - A buzzard (m = 9.29 kg) is flying in circular...Ch. 13 - An object of mass M isthrown with a velocity v0 at...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - A thin rod of length 2.65 m and mass 13.7 kg is...Ch. 13 - Prob. 47PQCh. 13 - Two particles of mass m1 = 2.00 kgand m2 = 5.00 kg...Ch. 13 - A turntable (disk) of radius r = 26.0 cm and...Ch. 13 - CHECK and THINK Our results give us a way to think...Ch. 13 - Prob. 51PQCh. 13 - Prob. 52PQCh. 13 - Two children (m = 30.0 kg each) stand opposite...Ch. 13 - A disk of mass m1 is rotating freely with constant...Ch. 13 - Prob. 55PQCh. 13 - Prob. 56PQCh. 13 - The angular momentum of a sphere is given by...Ch. 13 - Prob. 58PQCh. 13 - Prob. 59PQCh. 13 - Prob. 60PQCh. 13 - Prob. 61PQCh. 13 - Prob. 62PQCh. 13 - A uniform cylinder of radius r = 10.0 cm and mass...Ch. 13 - Prob. 64PQCh. 13 - A thin, spherical shell of mass m and radius R...Ch. 13 - To give a pet hamster exercise, some people put...Ch. 13 - Prob. 67PQCh. 13 - Prob. 68PQCh. 13 - The velocity of a particle of mass m = 2.00 kg is...Ch. 13 - A ball of mass M = 5.00 kg and radius r = 5.00 cm...Ch. 13 - A long, thin rod of mass m = 5.00 kg and length =...Ch. 13 - A solid sphere and a hollow cylinder of the same...Ch. 13 - A uniform disk of mass m = 10.0 kg and radius r =...Ch. 13 - When a person jumps off a diving platform, she...Ch. 13 - One end of a massless rigid rod of length is...Ch. 13 - A uniform solid sphere of mass m and radius r is...Ch. 13 - Prob. 77PQCh. 13 - A cam of mass M is in the shape of a circular disk...Ch. 13 - Prob. 79PQCh. 13 - Consider the downhill race in Example 13.9 (page...Ch. 13 - Prob. 81PQ
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- A suspicious physics student watches a stunt performed at an ice show. In the stunt, a performer shoots an arrow into a bale of hay (Fig. P11.24). Another performer rides on the bale of hay like a cowboy. After the arrow enters the bale, the balearrow system slides roughly 5 m along the ice. Estimate the initial speed of the arrow. Is there a trick to this stunt? FIGURE P11.24arrow_forwardEstimate the kinetic energy of the following: a. An ant walking across the kitchen floor b. A baseball thrown by a professional pitcher c. A car on the highway d. A large truck on the highwayarrow_forwardIn each situation shown in Figure P8.12, a ball moves from point A to point B. Use the following data to find the change in the gravitational potential energy in each case. You can assume that the radius of the ball is negligible. a. h = 1.35 m, = 25, and m = 0.65 kg b. R = 33.5 m and m = 756 kg c. R = 33.5 m and m = 756 kg FIGURE P8.12 Problems 12, 13, and 14.arrow_forward
- A small particle of mass m is pulled to the top of a friction less half-cylinder (of radius R) by a light cord that passes over the top of the cylinder as illustrated in Figure P7.15. (a) Assuming the particle moves at a constant speed, show that F = mg cos . Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (b) By directly integrating W=Fdr, find the work done in moving the particle at constant speed from the bottom to the top of the hall-cylinder. Figure P7.15arrow_forwardA small block of mass m = 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point relative to point . (b) the kinetic energy of the block at point , (c) its speed at point , and (d) its kinetic energy and the potential energy when the block is at point . Figure P7.45 Problems 45 and 46.arrow_forwardA block is placed on top of a vertical spring, and the spring compresses. Figure P8.24 depicts a moment in time when the spring is compressed by an amount h. a. To calculate the change in the gravitational and elastic potential energies, what must be included in the system? b. Find an expression for the change in the systems potential energy in terms of the parameters shown in Figure P8.24. c. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy when the blocks displacement is h = 0.0650 m, relative to its initial position. FIGURE P8.24arrow_forward
- A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P8.43). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point relative to point . (b) the kinetic energy of the block at point . (c) its speed at point B, and (d) its kinetic energy and the potential energy when the block is at point . Figure P8.43 Problems 43 and 44.arrow_forwardA light spring is attached to a block of mass 4m at rest on a frictionless, horizontal table. A second block of mass m is now placed on the table, in contact with the free end of the spring, and the two blocks are pushed together (Fig. P10.78). When the blocks are released, the more massive block moves to the left at 2.50 m/s. a. What is the speed of the less massive block? b. If m = 1.00 kg, what is the elastic potential energy of the system before it is released from rest? FIGURE P10.78arrow_forwardReview. A uniform board of length L is sliding along a smooth, frictionless, horizontal plane as shown in Figure P8.45a. The board then slides across the boundary with a rough horizontal surface. The coefficient of kinetic friction between the board and the second surface is k. (a) Find the acceleration of the board at the moment its front end has traveled a distance x beyond the boundary. (b) The board stops at the moment its back end reaches the boundary as shown in Figure P8.45b. Find the initial speed v of the board. Figure P8.45arrow_forward
- A small object is attached to two springs of the same length l, but with different spring constants k1 and k2 as shown in Figure P9.31. Initially, both springs are relaxed. The object is then displaced straight along the x axis from xi to xf. Find an expression for the work done by the springs on the object.arrow_forwardA particle moves in the xy plane (Fig. P9.30) from the origin to a point having coordinates x = 7.00 m and y = 4.00 m under the influence of a force given by F=3y2+x. a. What is the work done on the particle by the force F if it moves along path 1 (shown in red)? b. What is the work done on the particle by the force F if it moves along path 2 (shown in blue)? c. What is the work done on the particle by the force F if it moves along path 3 (shown in green)? d. Is the force F conservative or nonconservative? Explain. FIGURE P9.30 In each case, the work is found using the integral of Fdr along the path (Equation 9.21). W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz) (a) The work done along path 1, we first need to integrate along dr=dxi from (0,0) to (7,0) and then along dr=dyj from (7,0) to (7,4): W1=x=0;y=0x=7;y=0(3y2i+xj)(dxi)+x=7;y=0x=7;y=4(3y2i+xj)(dyj) Performing the dot products, we get W1=x=0;y=0x=7;y=03y2dx+x=7;y=0x=7;y=4xdy Along the first part of this path, y = 0 therefore the first integral equals zero. For the second integral, x is constant and can be pulled out of the integral, and we can evaluate dy. W1=0+x=7;y=0x=7;y=4xdy=xy|x=7;y=0x=7;y=4=28J (b) The work done along path 2 is along dr=dyj from (0,0) to (0,4) and then along dr=dxi from (0,4) to (7,4): W2=x=0;y=0x=0;y=4(3y2i+xj)(dyj)+x=0;y=4x=7;y=4(3y2i+xj)(dyi) Performing the dot product, we get: W2=x=0;y=0x=0;y=4xdy+x=0;y=4x=7;y=43y2dx Along the first part of this path, x = 0. Therefore, the first integral equals zero. For the second integral, y is constant and can be pulled out of the integral, and we can evaluate dx. W2=0+3y2x|x=0;y=4x=7;y=4=336J (c) To find the work along the third path, we first write the expression for the work integral. W=rtrfFdr=rtrf(Fxdx+Fydy+Fzdz)W=rtrf(3y2dx+xdy)(1) At first glance, this appears quite simple, but we cant integrate xdy=xy like we might have above because the value of x changes as we vary y (i.e., x is a function of y.) [In parts (a) and (b), on a straight horizontal or vertical line, only x or y changes]. One approach is to parameterize both x and y as a function of another variable, say t, and write each integral in terms of only x or y. Constraining dr to be along the desired line, we can relate dx and dy: tan=dydxdy=tandxanddx=dytan(2) Now, use equation (2) in (1) to express each integral in terms of only one variable. W=x=0;y=0x=7;y=43y2dx+x=0;y=0x=7;y=4xdyW=y=0y=43y2dytan+x=0x=7xtandx We can determine the tangent of the angle, which is constant (the angle is the angle of the line with respect to the horizontal). tan=4.007.00=0.570 Insert the value of the tangent and solve the integrals. W=30.570y33|y=0y=4+0.570x22|x=0x=7W=112+14=126J (d) Since the work done is not path-independent, this is non-conservative force. Figure P9.30ANSarrow_forwardA jack-in-the-box is actually a system that consists of an object attached to the top of a vertical spring (Fig. P8.50). a. Sketch the energy graph for the potential energy and the total energy of the springobject system as a function of compression distance x from x = xmax to x = 0, where xmax is the maximum amount of compression of the spring. Ignore the change in gravitational potential energy. b. Sketch the kinetic energy of the system between these points the two distances in part (a)on the same graph (using a different color). FIGURE P8.50 Problems 50 and 79arrow_forward
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