Hint 1A small block of mass 246 g starts at rest at A, slides to B where its speed is vg = 5.4 m/s, then slidesalong the horizontal surface a distance 10 m before coming to rest at C.A4.0 mB-10 m-Image Descriptiona. What is the work of friction along the curved surface?W;Jb. What is the coefficient of kinetic friction along the horizontal surface?Submit Question

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Answered: A small block of mass 246 g starts at…

A small block of mass 246 g starts at rest at A, slides to B where its speed is vg = 5.4 m/s, then slides along the horizontal surface a distance 10 m before coming to rest at C. A 4.0 m B 10 m

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter5: Energy

Section: Chapter Questions

Problem 40P: (a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force...

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(a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force F at an angle with respect to the horizontal. The coefficient of kinetic friction between block and table is k the force exerted by friction equal to kmg? If not, what is the force exerted by friction? (b) How much work is done by the friction force and by F? (Dont forget the signs.) (c) Identify all the forces that do no work on the block, (d) Let m = 2.00 kg, x = 4.00 m, = 37.0, F= 15.0 N, and k = 0.400, and find I the answers to parts (a) and (b). Figure P5.39
(a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force F at an angle with respect to the horizontal. The coefficient of kinetic friction between block and table is k the force exerted by friction equal to kmg? If not, what is the force exerted by friction? (b) How much work is done by the friction force and by F? (Dont forget the signs.) (c) Identify all the forces that do no work on the block, (d) Let m = 2.00 kg, x = 4.00 m, = 37.0, F= 15.0 N, and k = 0.400, and find I the answers to parts (a) and (b). Figure P5.39
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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
A force F(x)=(5.0x2+7.0x)N acts on a particle as it moves along the positive x-axis. (a)How much work does the force do on the particle as it moves from x2.0 to x=5.0 m? (b) Picking a convenient reference point of the potential energy to be zero at x=, find the potential energy for this force.