The system in the figure is released from rest. The coefficient of kinetic friction between block A and the table is 0.200. If block A moves to the right by 0.500 m, what is the work done by the weight of block B? N mA=3.00 kg fk mg=2.00 kg O 14.7 J O 19.6 J O4.9 J 9.8 J A 29.4 N 19.6 N

The system in the figure is released from rest. The coefficient of kinetic friction between block A and the table is 0.200. If block A moves to the right by 0.500 m, what is the work done by the weight of block B? N mA=3.00 kg fk mg=2.00 kg O 14.7 J O 19.6 J O4.9 J 9.8 J A 29.4 N 19.6 N

Name: The system in the figure is released from rest. The coefficient of ki
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Description: The system in the figure is released from rest. The coefficient of kinetic friction between block A and the table is 0.200. If block A moves to the right by 0.500 m, what is the work done by the weight of block B?NMA=3.00 kgfkB mg=2.00 kgO 14.7 JO 1

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Conservation Of Energy

Section: Chapter Questions

Problem 3P: A block of mass m = 5.00 kg is released from point and slides on the frictionless track shown in...

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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 P9.30ANS
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Figure P9.65A shows a crate attached to a rope that is extended over an ideal pulley. Boris pulls on the other end of the rope with a constant force until the crate has risen a total distance of 6.53 m (Fig. P9.65B). If the crate has a mass of 81.36 kg, what is the average power exerted by Boris, assuming he accomplishes the task in 5.33 s? FIGURE P9.65
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.
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