Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 43, Problem 62AP
(a)
To determine
The equilibrium position of the
(b)
To determine
The depth
(c)
To determine
The maximum force toward the negative
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Chapter 43 Solutions
Physics for Scientists and Engineers with Modern Physics, Technology Update
Ch. 43.1 - For each of the following atoms or molecules,...Ch. 43.2 - Prob. 43.2QQCh. 43.2 - Prob. 43.3QQCh. 43 - Prob. 1OQCh. 43 - Prob. 2OQCh. 43 - Prob. 3OQCh. 43 - Prob. 4OQCh. 43 - Prob. 5OQCh. 43 - Prob. 6OQCh. 43 - Prob. 7OQ
Ch. 43 - Prob. 1CQCh. 43 - Prob. 2CQCh. 43 - Prob. 3CQCh. 43 - Prob. 4CQCh. 43 - Prob. 5CQCh. 43 - Prob. 6CQCh. 43 - Prob. 7CQCh. 43 - Prob. 8CQCh. 43 - Discuss models for the different types of bonds...Ch. 43 - Prob. 10CQCh. 43 - Prob. 1PCh. 43 - Prob. 2PCh. 43 - Prob. 3PCh. 43 - Prob. 4PCh. 43 - Prob. 5PCh. 43 - Prob. 6PCh. 43 - Prob. 7PCh. 43 - Prob. 8PCh. 43 - Prob. 9PCh. 43 - Prob. 10PCh. 43 - Prob. 12PCh. 43 - Prob. 13PCh. 43 - Prob. 14PCh. 43 - Prob. 15PCh. 43 - Prob. 16PCh. 43 - The nuclei of the O2 molecule are separated by a...Ch. 43 - Prob. 18PCh. 43 - Prob. 19PCh. 43 - Prob. 20PCh. 43 - Prob. 21PCh. 43 - Prob. 22PCh. 43 - Prob. 23PCh. 43 - Prob. 24PCh. 43 - Prob. 25PCh. 43 - Prob. 27PCh. 43 - Prob. 28PCh. 43 - Prob. 29PCh. 43 - Prob. 30PCh. 43 - Prob. 31PCh. 43 - Prob. 32PCh. 43 - Prob. 33PCh. 43 - Prob. 34PCh. 43 - Prob. 35PCh. 43 - Prob. 36PCh. 43 - Prob. 37PCh. 43 - Prob. 38PCh. 43 - Prob. 39PCh. 43 - Prob. 40PCh. 43 - Prob. 41PCh. 43 - Prob. 42PCh. 43 - Prob. 43PCh. 43 - Prob. 44PCh. 43 - Prob. 45PCh. 43 - Prob. 46PCh. 43 - Prob. 47PCh. 43 - Prob. 49PCh. 43 - Prob. 50PCh. 43 - Prob. 51PCh. 43 - A direct and relatively simple demonstration of...Ch. 43 - Prob. 53PCh. 43 - Prob. 54APCh. 43 - Prob. 55APCh. 43 - Prob. 56APCh. 43 - Prob. 57APCh. 43 - Prob. 58APCh. 43 - Prob. 59APCh. 43 - Prob. 61APCh. 43 - Prob. 62APCh. 43 - Prob. 63CPCh. 43 - As an alternative to Equation 43.1, another useful...
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- 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 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 train moves along the tracks at a constant speed u. A woman on the train throws a ball of mass m straight ahead with a speed υ with respect to herself. (a) What is the kinetic energy gain of the ball as measured by a person on the train? (b) by a person standing by the railroad track? (c) How much work is done by the woman throwing he ball and (d) by the train?arrow_forward
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