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 23PQ
A constant force of magnitude 4.75 N is exerted on an object. The force’s direction is 60.0° counterclockwise from the positive x axis in the xy plane, and the object’s displacement is
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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 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_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 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_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
- An object of mass m = 5.8 kg moves under the influence of one force. That force causes the object to move along a path given by x = 6.0 + 5.0t + 2.0t2, where x is in meters and t is in seconds. Calculate the work done by the force on the object from t = 2.0 s to t = 7.0 s.arrow_forward(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?arrow_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_forward
- 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 P7.31arrow_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_forwardThe force acting on a particle is Fx = (8x 16), where F is in newtons anti x is in meters. (a) Make a plot of this force versus x from x = 0 to x = 3.00 m. (b) From your graph, find the net work done by this force on the particle as it moves from x = 0 to x = 3.00 m.arrow_forward
- A 4.00-kg particle moves from the origin to position ©, having coordinates x = 5.00 m and y = 5.00 m (Fig. P6.42). One force on the particle is the gravitational force acting in the negative y direction. Using Equation 6.3, 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 P6.42 Problems 42 through 45.arrow_forwardA 4.00-kg particle moves along the x axis. Its position O varies with time according to x = t + 2.0t3, where x is in meters and t is in seconds. Find (a) the kinetic energy of the particle at any time t (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t and (d) the work done on the particle in the interval t = 0 to t = 2.00 s.arrow_forwardRepeat 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_forward
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