Chapter 07, Problem 061PRINTER VERSION1 BACKNEXYour answer is partially correct. Try again.What work is done by a force F = (8.00XN)î + (5.00N),with x in meters, that moves a particle from a position T = (5.00 m)i + (3.00 m)j7;=-(5.00 m)î – (3.00 m)}?to a positionNumberUnitsthe tolerance is +/-2%Click if you would like to Show Work for this question:Open Show WorkQuestion Attempts: UnlimitedSAVE FOR LATERSUBMIT ANSWERsearch9:08 PMENG4/4/2021pausebreakprt scinsertdelet13)17home10 endPgupback35.8.近

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Description: Chapter 07, Problem 061PRINTER VERSION1 BACKNEXYour answer is partially correct. Try again.What work is done by a force F = (8.00XN)î + (5.00N),with x in meters, that moves a particle from a position T = (5.00 m)i + (3.00 m)j7;=-(5.00 m)î – (3.00 m)

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PRINTER VERSION 1 BACK Chapter 07, Problem 061 Your answer is partially correct. Try again. What work is done by a force F = (8.00 x N)î + (5.00 N )j, 7;=-(5.00 m)î – (3.00 m)ĵ? ,with x in meters, that moves a particle from a position T = (5.00m)i+ (3.00 m) to a position %3D Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work SAVE FOR LATER SUBMIT ANSWER Question Attempts: Unlimited

PRINTER VERSION 1 BACK Chapter 07, Problem 061 Your answer is partially correct. Try again. What work is done by a force F = (8.00 x N)î + (5.00 N )j, 7;=-(5.00 m)î – (3.00 m)ĵ? ,with x in meters, that moves a particle from a position T = (5.00m)i+ (3.00 m) to a position %3D Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work SAVE FOR LATER SUBMIT ANSWER Question Attempts: Unlimited

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 6OQ: As a simple pendulum swings back and forth, the forces acting on the suspended object are (a) the...

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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.00 slope at constant speed, as shown below. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by fiction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0° below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?
In a Coyote/Road Runner cartoon clip (https://openstaxcollege.org/l/21coyroadcarcl), a spring expands quickly and sends the coyote into a rock. If the spring extended 5 m and sent the coyote of mass 20 kg to a speed of 15 m/s, (a) what is the spring constant of this spring? (b) If the coyote were sent vertically into the air with the energy given to him by the spring, how high could he go if there were no non-conservative forces?
A person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people ‘would it take to nm a 4.O0-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW?
“ E=K+Uconstant is a special case of the work energy theorem.” Discuss this statement.
A large cruise ship of mass 6.50 107 kg has a speed of 12.0 m/s at some instant. (a) What is the ships kinetic energy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.50 km?
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
As a simple pendulum swings back and forth, the forces acting on the suspended object are (a) the gravitational force, (b) the tension in the supporting cord, and (c) air resistance. (i) Which of these forces, if any, does no work on the pendulum at any time? (ii) Which of these forces does negative work on the pendulum at all times during its motion?
The 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.