Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Question
Chapter 6, Problem 122P
(a)
To determine
The potential energy at
(b)
To determine
Particle’s total, potential and kinetic energies at
(c)
To determine
Explain the motion of the particle, whether the particle ever turns around and start moving to the right, the value of x where the speed is greatest.
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Check out a sample textbook solutionChapter 6 Solutions
Physics
Ch. 6.2 - Prob. 6.2CPCh. 6.2 - Prob. 6.1PPCh. 6.2 - Prob. 6.2PPCh. 6.3 - Prob. 6.3PPCh. 6.3 - Prob. 6.4PPCh. 6.3 - Prob. 6.3ACPCh. 6.3 - Prob. 6.3BCPCh. 6.4 - Prob. 6.4ACPCh. 6.4 - Prob. 6.4BCPCh. 6.4 - Prob. 6.5PP
Ch. 6.4 - Prob. 6.6PPCh. 6.5 - Prob. 6.5CPCh. 6.5 - Prob. 6.7PPCh. 6.5 - Prob. 6.8PPCh. 6.6 - Prob. 6.9PPCh. 6.6 - Prob. 6.10PPCh. 6.7 - Prob. 6.7CPCh. 6.7 - Practice Problem 6.11 A Misfire
The same dart gun...Ch. 6.7 - Prob. 6.12PPCh. 6.8 - Prob. 6.13PPCh. 6.8 - Prob. 6.14PPCh. 6 - Prob. 1CQCh. 6 - Prob. 2CQCh. 6 - Prob. 3CQCh. 6 - Prob. 4CQCh. 6 - Prob. 5CQCh. 6 - Prob. 6CQCh. 6 - Prob. 7CQCh. 6 - Prob. 8CQCh. 6 - Prob. 9CQCh. 6 - Prob. 10CQCh. 6 - Prob. 11CQCh. 6 - Prob. 12CQCh. 6 - Prob. 13CQCh. 6 - Prob. 1MCQCh. 6 - 2. If a kangaroo on Earth can jump from a standing...Ch. 6 - Prob. 3MCQCh. 6 - Prob. 4MCQCh. 6 - Prob. 5MCQCh. 6 - Prob. 6MCQCh. 6 - Prob. 7MCQCh. 6 - Prob. 8MCQCh. 6 - Prob. 9MCQCh. 6 - Questions 9 and 10. A simple catapult, consisting...Ch. 6 - Prob. 11MCQCh. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 4PCh. 6 - Prob. 5PCh. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10PCh. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - Prob. 22PCh. 6 - Prob. 23PCh. 6 - Prob. 24PCh. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Prob. 27PCh. 6 - Prob. 28PCh. 6 - Problems 29–32. A skier passes through points A–E...Ch. 6 - Prob. 30PCh. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Prob. 34PCh. 6 - 35. Emil is tossing an orange of mass 0.30 kg into...Ch. 6 - Prob. 36PCh. 6 - 37. An arrangement of two pulleys, as shown in the...Ch. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - 47. Refer to Problems 11-14. Find conservation of...Ch. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - 53. What is the minimum speed with which a meteor...Ch. 6 - 54. A projectile with mass of 500 kg is launched...Ch. 6 - Prob. 55PCh. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - Prob. 71PCh. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 76PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - Prob. 86PCh. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - Prob. 94PCh. 6 - Prob. 95PCh. 6 - Prob. 97PCh. 6 - Prob. 96PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 101PCh. 6 - Prob. 100PCh. 6 - Prob. 102PCh. 6 - Prob. 103PCh. 6 - Prob. 104PCh. 6 - Prob. 105PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - Prob. 110PCh. 6 - Prob. 111PCh. 6 - Prob. 112PCh. 6 - Prob. 106PCh. 6 - Prob. 113PCh. 6 - Prob. 114PCh. 6 - Prob. 115PCh. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - Prob. 118PCh. 6 - Prob. 119PCh. 6 - Prob. 120PCh. 6 - Prob. 130PCh. 6 - Prob. 125PCh. 6 - Problems 121 and 122.A particle is constrained to...Ch. 6 - Prob. 122PCh. 6 - Prob. 123PCh. 6 - Prob. 124PCh. 6 - Prob. 132PCh. 6 - Prob. 126PCh. 6 - Prob. 127PCh. 6 - Prob. 128PCh. 6 - Prob. 129PCh. 6 - Prob. 131PCh. 6 - Prob. 142PCh. 6 - Prob. 140PCh. 6 - Prob. 133PCh. 6 - Prob. 134PCh. 6 - Prob. 136PCh. 6 - Prob. 137PCh. 6 - Prob. 138PCh. 6 - Prob. 135PCh. 6 - Prob. 139PCh. 6 - Prob. 141P
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- Physics Review A team of huskies performs 7 440 J of work on a loaded sled of mass 124 kg, drawing it from rest up a 4.60-m high snow-covered rise while the sled loses 1 520 J due to friction, (a) What is the net work done on the sled by the huskies and friction? (b) What is the change in the sleds potential energy? (c) What is the speed of the sled at the top of the rise? (See Section 5.5.)arrow_forward(a) Can the kinetic energy of a system be negative? (b) Can the gravitational potential energy of a system be negative? Explain.arrow_forwardA particle moves in one dimension under the action of a conservative force. The potential energy of the system is given by the graph in Figure P8.55. Suppose the particle is given a total energy E, which is shown as a horizontal line on the graph. a. Sketch bar charts of the kinetic and potential energies at points x = 0, x = x1, and x = x2. b. At which location is the particle moving the fastest? c. What can be said about the speed of the particle at x = x3? FIGURE P8.55arrow_forward
- 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?arrow_forwardAs a simple pendulum swings back and forth, the forces acting on the suspended object are the force of gravity, the tension in the supporting cord, and air resistance, (a) Which of these forces, if any, does no work on the pendulum? (b) Which of these forces does negative work at all times during the pendulums motion? (c) Describe the work done by the force of gravity while the pendulum is swinging.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_forward
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