Concept explainers
You have a new internship, where you are helping to design a new freight yard for the train station in your city. There will be a number of dead-end sidings where single cars can be stored until they are needed. To keep the cars from running off the tracks at the end of the siding, you have designed a combination of two coiled springs as illustrated in Figure P7.41. When a car moves to the right in the figure and strikes the springs, they exert a force to the left on the car to slow it down.
Both springs are described by Hooke’s law and have spring constants k1 = 1 600 N/m and k2 = 3 400 N/m. After the first spring compresses by a distance of d = 30.0 cm, the second spring acts with the first to increase the force to the left on the car in Figure P7.41. When the spring with spring constant k2 compresses by 50.0 cm, the coils of both springs are pressed together, so that the springs can no longer compress. A typical car on the siding has a mass of 6 000 kg. When you present your design to your supervisor, he asks you for the maximum speed that a car can have and be stopped by your device.
Figure P7.41
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Chapter 7 Solutions
Physics for Scientists and Engineers
- A nonconstant force is exerted on a particle as it moves in the positive direction along the x axis. Figure P9.26 shows a graph of this force Fx versus the particles position x. Find the work done by this force on the particle as the particle moves as follows. a. From xi = 0 to xf = 10.0 m b. From xi = 10.0 to xf = 20.0 m c. From xi = 0 to xf = 20.0 m FIGURE P9.26 Problems 26 and 27.arrow_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 particle moves in one dimension under the action of a conservative force. The potential energy of the system is given by the graph in Figure P8.55. Suppose the particle is given a total energy E, which is shown as a horizontal line on the graph. a. Sketch bar charts of the kinetic and potential energies at points x = 0, x = x1, and x = x2. b. At which location is the particle moving the fastest? c. What can be said about the speed of the particle at x = x3? FIGURE P8.55arrow_forwardTo give a pet hamster exercise, some people put the hamster in a ventilated ball andallow it roam around the house(Fig. P13.66). When a hamsteris in such a ball, it can cross atypical room in a few minutes.Estimate the total kinetic energyin the ball-hamster system. FIGURE P13.66 Problems 66 and 67arrow_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_forward
- The motion of a box of mass m = 2.00 kg along the x axis can be described by the function x = 4.00 + 3.00t2+ 2.00t3, where x is in meters and t is in seconds. a. What is the kinetic energy of the box as a function of time? b. What are the acceleration of the box and the force acting on the box as a function of time? c. What is the power delivered to the box as a function of time? d. What is the work performed on the particle during the time interval t = 1.00 s to t = 3.00 s?arrow_forwardA small 0.65-kg box is launched from rest by a horizontal spring as shown in Figure P9.50. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. Kinetic friction between the box and the track is negligible on the hill, but the coefficient of kinetic friction between the box and the horizontal parts of track is 0.35. The spring has a spring constant of 34.5 N/m, and is compressed 30.0 cm with the box attached. The block remains on the track at all times. a. What would you include in the system? Explain your choice. b. Calculate d.arrow_forwardA block is hung from a vertical spring. The spring stretches (h = 0.0650 m) as shown for a particular instant in time in Figure P8.26. Consider the Earth, spring, and block to be in the system. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy between the two times depicted in the figure. FIGURE P8.26arrow_forward
- A roller-coaster car shown in Figure P7.82 is released from rest from a height h and then moves freely with negligible friction. The roller-coaster track includes a circular loop of radius R in a vertical plane. (a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop in terms of R. (b) Now assume the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the cars weight. The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. Figure P5.22 (page 149) shows an actual design.arrow_forwardIn a basketball game, lebron made a dunk. At the time he was hanging on to the rim, the rim was displaced 15cm and he was 1.5m above the floor. Let us say that lebron has a point mass of 110 kilograms, and the force constant of the rim is 7 kN/m what is his total potential energy while he is hanging on the rim? A. 79 J B. 80 J C. 2405 J D. 1617 J E. 1696 Jarrow_forwardConsider a fruit hanging on a twig with potential energy of 150 J. If the fruit suddenly falls to the ground, what is the kinetic energy of the fruit as it lands?a. 150.0 Jb. 0 Jc. 1470 Jd. 159.8 Jarrow_forward
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