A force F = (4î - 31) = (4 1 - 3 ĵ) N acts on a particle that undergoes a displacement Ař = (2 î + ĵ) m. (a) Find the work done by the force on the particle. J (b) What is the angle between F and Ar? O

A force F = (4î - 31) = (4 1 - 3 ĵ) N acts on a particle that undergoes a displacement Ař = (2 î + ĵ) m. (a) Find the work done by the force on the particle. J (b) What is the angle between F and Ar? O

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter7: Energy Of A System

Section: Chapter Questions

Problem 7.11P: A force F = (6i 2j) N acts on a panicle that under-goes a displacement r = (3i + j) m. Find (a) the...

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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
Consider 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.
A mysterious force acts on all particles along a particular line and always points towards a particular point P on the line. The magnitude of the force on a particle increases as the cube of the distance from that point; that is Fr3 , if the distance from P to the position of the particle is r. Let b be the proportionality constant, and write the magnitude of the force as F=br3. Find the potential energy of a particle subjected to this force when the particle is at a distance D from P, assuming the potential energy to be zero when the particle is at P.
(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?
A 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 P7.31
If 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.
A 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.
(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.
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 ?
Consider a particle on which several forces act, one of which is known to be constant in time: . As a result, the particle moves along the x-axis from x=0 to x=5 m in some time interval. What is the work done by ?
A particle moves along the xaxis from x = 12.8 m to x = 23.7 m under the influence of a force F=375x3+3.75x where F is in newtons and x is in meters. Using numerical integration, determine the work done by this force on the particle during this displacement. Your result should he accurate to within 2%.
A particle moves along a curved path y(x)=(10m){1+cos[0.1m1]x} , from x=0 to x=10m , subject to a tangential force of variable magnitude F(x)=(10N)sin[(0.1m-1)x] . How much work does the force do? (Hint: Consult a table of integrals or use a numerical integration program.)