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A particle moves along a straight path through displacement
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- A block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle = 25.0 below the horizontal as shown in Figure P6.3. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, (c) the gravitational force, and (d) the net force on the block. Figure P6.3arrow_forwardConsider a particle on which a force acts that depends on the position of the particle. This force is given by . Find the work done by this force when the particle moves from the origin to a point 5 meters to the right on the x-axis.arrow_forwardAs a young man, Tarzan climbed up a vine to reach his tree house. As he got older, he decided to build and use a staircase instead. Since the work of the gravitational force mg is path Independent, what did the King of the Apes gain in using stairs?arrow_forward
- Repeat the preceding problem, but this time, suppose that the work done by air resistance cannot be ignored. Let the work done by the air resistance when the skier goes from A to B along the given hilly path be —2000 J. The work done by air resistance is negative since the air resistance acts in the opposite direction to the displacement. Supposing the mass of the skier is 50 kg, what is the speed of the skier at point B ?arrow_forwardA block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless horizontal table by a constant applied force of magnitude F = 16.0 N directed at an angle = 25.0 below the horizontal as shown in Figure P5.8. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block. Figure P5.8arrow_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_forward
- In Figure 5.5 (a)-(d), a block moves to the right in the positive x-direction through the displacement x while under the influence of a force with the same magnitude F. Which of the following is the correct order of the amount of work done by the force F, from most positive to most negative? (a) d, c, a, b (b) c, a, b, d (c) c, a, d, barrow_forwardA block of mass m = 2.50 kg is pushed a distance d = 2.20 m along a frictionless, horizontal table by a constant applied force of magnitude F = 16.0 N directed at ail angle = 25 below the horizontal as shown in Figure P7.5. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, (c) the gravitational force, and (d) the net force on the block.arrow_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
- The force acting on a panicle varies as shown in Figure la P7.14. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m. (b) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 to x = 10.0 m.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_forwardThe force acting on a particle varies as shown in Figure P6.14. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x= 10.0 m, and (c) from x = 0 to x = 10.0 m.arrow_forward
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