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CALC In an experiment, one of the forces exerted on a proton is
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- (a) A force F=(4xi+3yj), where F is in newtons and x and y are in meters, acts on an object as the object moves in the x direction from the origin to x = 5.00 m. Find the work W=Fdr done by the force on the object. (b) What If? Find the work W=Fdr done by the force on the object if it moves from the origin to (5.00 m, 5.00 m) along a straightline path making an angle of 45.0 with the positive x axis. Is the work done by this force dependent on the path taken between the initial and final points?arrow_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 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_forward
- A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance?arrow_forwardA 4.00-kg particle moves from the origin to position , having coordinates x = 5.00 m and y = 5.00 m (Fig. P7.31). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 7.3, 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 P7.31arrow_forward(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.arrow_forward
- A particle of mass 2.0 kg moves under the influence of the force F(x)=(3/x)N. If its speed at x=2.0 m is v=6.0 m/s, what is its speed at x = 7.0 m?arrow_forwardPlease Asaparrow_forwardTwo persons were practicing social distancing to alleviate the spread of COVID-19. Social distancing requires a minimum of 3 feet distance from person to person. However, due to their eagerness to share some rumors, the two persons decided to approach each other and now within the intimate distance (12 inches). If person A weighs 60 kg. and person B weighs 68 kg., determine the work done for this act. Take G = 6.67408 × 10-11 m³ kg-ls-2. A. 6.8076 ergs C. 5.9559 ergs D. 5.5599 ergs B. 8.0766 ergsarrow_forward
- Two persons were practicing social distancing to alleviate the spread of COVID-19. Social distancing requires a minimum of 3 feet distance from person to person. However, due to their eagerness to share some rumors, the two persons decided to approach each other and now within the intimate distance (12 inches). If person A weighs 60 kg. and person B weighs 68 kg., determine the work done for this act. Take G = 6.67408 × 10-11 m3 kg-1s-2arrow_forwardProblem A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of Kq? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that K1=(1/2)mv², K2=(1/2)mv2 But it was mentioned that: so, Kz in terms of v is K2=( )mv²1 Substituting the expression for K1 results to K2=( Since work done is W=AK=K -K Evaluating results to W=( )K1arrow_forwardA body with mass 0.40 kg moves in one dimension according to a potential of theform U(x) = A/x3-B/x , where A = 4.0 Jm3 abd B = 3.0 Jm, and we take x > 0. (a) If the speed of the body at x = 4.0m is v = 3.9 ms−1, what is the speed at x = 1.5 m? (b) Explain briefly what type(s) of motion are possible in this potential if the totalenergy E = 3.0 J.arrow_forward
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