Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 9, Problem 33PQ
To determine
Prove that the work done by universal gravity in a closed path is zero.
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Chapter 9 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 9.4 - In the three cases shown in Figure 9.11, a force...Ch. 9.6 - Prob. 9.2CECh. 9.6 - Prob. 9.3CECh. 9.7 - Prob. 9.4CECh. 9.7 - Prob. 9.5CECh. 9.9 - Prob. 9.6CECh. 9 - Pick an isolated system for the following...Ch. 9 - Prob. 2PQCh. 9 - Prob. 3PQCh. 9 - Prob. 4PQ
Ch. 9 - Prob. 5PQCh. 9 - Prob. 6PQCh. 9 - Prob. 7PQCh. 9 - A 537-kg trailer is hitched to a truck. Find the...Ch. 9 - Prob. 9PQCh. 9 - A helicopter rescues a trapped person of mass m =...Ch. 9 - Prob. 11PQCh. 9 - An object is subject to a force F=(512i134j) N...Ch. 9 - Prob. 13PQCh. 9 - Prob. 14PQCh. 9 - Prob. 15PQCh. 9 - Prob. 16PQCh. 9 - Prob. 17PQCh. 9 - Prob. 18PQCh. 9 - Prob. 19PQCh. 9 - Prob. 20PQCh. 9 - Prob. 21PQCh. 9 - Prob. 22PQCh. 9 - A constant force of magnitude 4.75 N is exerted on...Ch. 9 - In three cases, a force acts on a particle, and...Ch. 9 - An object of mass m = 5.8 kg moves under the...Ch. 9 - A nonconstant force is exerted on a particle as it...Ch. 9 - Prob. 27PQCh. 9 - Prob. 28PQCh. 9 - Prob. 29PQCh. 9 - A particle moves in the xy plane (Fig. P9.30) from...Ch. 9 - A small object is attached to two springs of the...Ch. 9 - Prob. 32PQCh. 9 - Prob. 33PQCh. 9 - Prob. 34PQCh. 9 - Prob. 35PQCh. 9 - Prob. 36PQCh. 9 - Prob. 37PQCh. 9 - Prob. 38PQCh. 9 - A shopper weighs 3.00 kg of apples on a...Ch. 9 - Prob. 40PQCh. 9 - Prob. 41PQCh. 9 - Prob. 42PQCh. 9 - Prob. 43PQCh. 9 - Prob. 44PQCh. 9 - Prob. 45PQCh. 9 - Prob. 46PQCh. 9 - Prob. 47PQCh. 9 - Prob. 48PQCh. 9 - Prob. 49PQCh. 9 - A small 0.65-kg box is launched from rest by a...Ch. 9 - A small 0.65-kg box is launched from rest by a...Ch. 9 - A horizontal spring with force constant k = 625...Ch. 9 - A box of mass m = 2.00 kg is dropped from rest...Ch. 9 - Prob. 54PQCh. 9 - Return to Example 9.9 and use the result to find...Ch. 9 - Prob. 56PQCh. 9 - Crall and Whipple design a loop-the-loop track for...Ch. 9 - Prob. 58PQCh. 9 - Calculate the force required to pull a stuffed toy...Ch. 9 - Prob. 60PQCh. 9 - Prob. 61PQCh. 9 - Prob. 62PQCh. 9 - An elevator motor moves a car with six people...Ch. 9 - Prob. 64PQCh. 9 - Figure P9.65A shows a crate attached to a rope...Ch. 9 - Prob. 66PQCh. 9 - Prob. 67PQCh. 9 - Prob. 68PQCh. 9 - Prob. 69PQCh. 9 - Prob. 70PQCh. 9 - Prob. 71PQCh. 9 - Estimate the power required for a boxer to jump...Ch. 9 - Prob. 73PQCh. 9 - Prob. 74PQCh. 9 - Prob. 75PQCh. 9 - Prob. 76PQCh. 9 - Prob. 77PQCh. 9 - Prob. 78PQCh. 9 - Prob. 79PQCh. 9 - A block of mass m = 0.250 kg is pressed against a...Ch. 9 - On a movie set, an alien spacecraft is to be...Ch. 9 - Prob. 82PQCh. 9 - A spring-loaded toy gun is aimed vertically and...Ch. 9 - Prob. 84PQCh. 9 - The motion of a box of mass m = 2.00 kg along the...Ch. 9 - Prob. 86PQCh. 9 - Prob. 87PQCh. 9 - Prob. 88PQ
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- A train moves along the tracks at a constant speed u. A woman on the train throws a ball of mass m straight ahead with a speed υ with respect to herself. (a) What is the kinetic energy gain of the ball as measured by a person on the train? (b) by a person standing by the railroad track? (c) How much work is done by the woman throwing he ball and (d) by the train?arrow_forwardIn three cases, a force acts on a particle, and the particle is displaced from an initial position to a final position. Figure 9.11 (page 255) shows the position-versus-force graphs, indicating the initial and final positions of the particle in each case. Find the work done by the force on the particle and sketch the force and displacement vectors along with the appropriate axis in each case.arrow_forwardA helicopter rescues a trapped person of mass m = 65.0 kg from a flooded river by lifting the person vertically upward using a winch and rope. The person is pulled 12.0 m into the helicopter with a constant force that is 15% greater than the persons weight. a. Find the work done by each of the forces acting on the person. b. Assuming the survivor starts from rest, determine his speed upon reaching the helicopter.arrow_forward
- A constant force of magnitude 4.75 N is exerted on an object. The forces direction is 60.0 counterclockwise from the positive x axis in the xy plane, and the objects displacement is r=(4.22.1j+1.6k)m. Calculate the work done by this force.arrow_forwardAt the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height ymax above where it started. Using conservation of energy, find expressions for a. the balls initial speed in terms of the gravitational acceleration g and the maximum height ymax and b. the height of the ball when it has speed v in terms of its current height y, the gravitational acceleration g, and the maximum height ymax.arrow_forwardA ball of mass 0.40 kg hangs straight down on a string of length 15 cm. It is then swung upward, keeping the string taut, until the string makes an angle of 65 with respect to the vertical. Find the change in the gravitational potential energy of the Earth-ball system.arrow_forward
- A cat’s crinkle ball toy of mass 15 g is thrown straight up with an initial speed of 3 m/s. Assume in this problem that air drag is negligible. (a) What is the kinetic energy of the ball as it leaves the hand? (b) How much work is done by the gravitational force during the ball’s rise to its peak? (c) What is the change in the gravitational potential energy of the ball during the rise to its peak? (d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height? (e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? (f) What is the maximum height the ball reaches?arrow_forwardA side view of a half-pipe at a skateboard park is shown in Figure P8.51. Sketch a graph of the gravitational potential energy of the skateboarderEarth system as a function of position for a skateboarder who travels from the left side of the half-pipe to the right side. Let the leftmost point be where x = 0 and the lowest point in the half-pipe be where U = 0. FIGURE P8.51arrow_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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