Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 34AP
To determine
To determine: The reason for the given situation is impossible.
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Physics for Scientists and Engineers with Modern Physics
Ch. 8.1 - Consider a block sliding over a horizontal surface...Ch. 8.2 - A rock of mass m is dropped to the ground from a...Ch. 8.2 - Three identical balls are thrown from the top of a...Ch. 8.3 - You are traveling along a freeway at 65 mi/h. Your...Ch. 8 - Prob. 1PCh. 8 - A 20.0-kg cannonball is fired from a cannon with...Ch. 8 - A block of mass m = 5.00 kg is released from point...Ch. 8 - At 11:00 a.m, on September 7, 2001, more than one...Ch. 8 - A light, rigid rod is 77.0 cm long. Its top end is...Ch. 8 - Prob. 6P
Ch. 8 - A crate of mass 10.0 kg is pulled up a rough...Ch. 8 - A 40.0-kg box initially at rest is pushed 5.00 m...Ch. 8 - Prob. 9PCh. 8 - As shown in Figure P8.10, a green bead of mass 25...Ch. 8 - At time ti, the kinetic energy of a particle is...Ch. 8 - A 1.50-kg object is held 1.20 m above a relaxed...Ch. 8 - Prob. 13PCh. 8 - An 80.0-kg skydiver jumps out of a balloon at an...Ch. 8 - You have spent a long day skiing and are tired....Ch. 8 - The electric motor of a model train accelerates...Ch. 8 - An energy-efficient lightbulb, taking in 28.0 W of...Ch. 8 - An older-model car accelerates from 0 to speed v...Ch. 8 - Prob. 19PCh. 8 - There is a 5K event coming up in your town. While...Ch. 8 - Prob. 21PCh. 8 - Energy is conventionally measured in Calories as...Ch. 8 - A block of mass m = 200 g is released from rest at...Ch. 8 - Prob. 24APCh. 8 - Prob. 25APCh. 8 - Review. As shown in Figure P8.26, a light string...Ch. 8 - Consider the blockspringsurface system in part (B)...Ch. 8 - Why is the following situation impossible? A...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - As the driver steps on the gas pedal, a car of...Ch. 8 - As it plows a parking lot, a snowplow pushes an...Ch. 8 - Prob. 33APCh. 8 - Prob. 34APCh. 8 - A horizontal spring attached to a wall has a force...Ch. 8 - Prob. 36APCh. 8 - Prob. 37APCh. 8 - Review. Why is the following situation impossible?...Ch. 8 - Prob. 39APCh. 8 - A pendulum, comprising a light string of length L...Ch. 8 - Prob. 41APCh. 8 - Prob. 42APCh. 8 - Prob. 43APCh. 8 - Starting from rest, a 64.0-kg person bungee jumps...Ch. 8 - Prob. 45CPCh. 8 - A uniform chain of length 8.00 m initially lies...Ch. 8 - Prob. 47CP
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- A 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_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_forwardA 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_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_forwardIf the net work done by external forces on a particle is zero, which of the following statements about the particle must be true? (a) Its velocity is zero. (b) Its velocity is decreased. (c) Its velocity is unchanged. (d) Its speed is unchanged. (e) More information is needed.arrow_forwardA loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0 above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy has transferred out of the motor by work by the time the car moves off the end of the track, which is of length 1 250 m?arrow_forward
- 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_forwardA 4.00-kg particle moves from the origin to position ©, having coordinates x = 5.00 m and y = 5.00 m (Fig. P6.42). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 6.3, calculate the work done by the gravitational force on the particle as it goes from O to © along (a) the purple path, (b) the red path, and (c) the blue path. (d) Your results should all be identical. Why? Figure P6.42 Problems 42 through 45.arrow_forward
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Kinetic Energy and Potential Energy; Author: Professor Dave explains;https://www.youtube.com/watch?v=g7u6pIfUVy4;License: Standard YouTube License, CC-BY