Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (Chs 1-42) Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition)
4th Edition
ISBN: 9780133953145
Author: Randall D. Knight (Professor Emeritus)
Publisher: PEARSON
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Chapter 9, Problem 18EAP
The two ropes seen in FIGURE EX9.18 are used to lower a 255 kg piano 5.00 m from a second-story window to the ground. How much work is done by each of the three forces?
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Chapter 9 Solutions
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (Chs 1-42) Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition)
Ch. 9 - If a particle’s speed increases by a factor of 3,...Ch. 9 - Prob. 2CQCh. 9 - 3. An elevator held by a single cable is ascending...Ch. 9 - The rope in FIGURE Q9.4 pulls the box to the left...Ch. 9 - 5. A 0.2 kg plastic cart and a 20 kg lead cart...Ch. 9 - A particle moving to the left is slowed by a force...Ch. 9 - 7. A particle moves in a vertical plane along the...Ch. 9 - 8. You need to raise a heavy block by pulling it...Ch. 9 - 9. A ball on a string travels once around a circle...Ch. 9 - A sprinter accelerates from rest. Is the work done...
Ch. 9 - 11. A Spring has an unstretched length of 10cm. It...Ch. 9 - 12. The left end of a spring is attached to a...Ch. 9 - The driver of a car traveling at 60 mph slams on...Ch. 9 - Prob. 14CQCh. 9 - Which has the larger kinetic energy, a 10 g bullet...Ch. 9 - At what speed does a 1000 kg compact car have the...Ch. 9 - 3. A mother has four times the mass of her young...Ch. 9 - 4. A horizontal rope with 15 N tension drags a 25...Ch. 9 - 5. A 25 kg box sliding to the left across a...Ch. 9 - A 2.0 kg book is lying on a 0.75-m-high table. You...Ch. 9 - Prob. 7EAPCh. 9 - Prob. 8EAPCh. 9 - 9. You throw a 5.5 g coin straight down at 4.0 m/s...Ch. 9 - Prob. 10EAPCh. 9 - 12. Evaluate the dot product if
and .
and .
Ch. 9 - 12. Evaluate the dot product if
and .
and .
Ch. 9 - 13. What is the angle ? between vectors and in...Ch. 9 - Prob. 14EAPCh. 9 - Prob. 15EAPCh. 9 - 16. A 25 kg air compressor is dragged up a rough...Ch. 9 - Prob. 17EAPCh. 9 - The two ropes seen in FIGURE EX9.18 are used to...Ch. 9 - 19. The three ropes shown in the bird’s-eye view...Ch. 9 - Prob. 20EAPCh. 9 - Prob. 21EAPCh. 9 - Prob. 22EAPCh. 9 - A particle moving on the x-axis experiences a...Ch. 9 - Prob. 24EAPCh. 9 - A horizontal spring with spring constant 750 N/m...Ch. 9 - 26. A 35-cm-long vertical spring has one end fixed...Ch. 9 - A 10-cm-long spring is attached to the ceiling....Ch. 9 - A 60 kg student is standing atop a spring in an...Ch. 9 -
29. A 5.0 kg mass hanging from a spring scale is...Ch. 9 - A horizontal spring with spring constant 85 N/m...Ch. 9 - 31. One mole (6.02 × 1023 atoms) of helium atoms...Ch. 9 - 32. A 55 kg softball player slides into second...Ch. 9 - A baggage handler throws a 15 kg suitcase along...Ch. 9 -
34. An 8.0 kg crate is pulled 5.0 m up a 30°...Ch. 9 - Justin, with a mass of 30 kg, is going down an...Ch. 9 - Prob. 36EAPCh. 9 - Prob. 37EAPCh. 9 - 38. How much energy is consumed by (a) a 1.2 kW...Ch. 9 - 39. At midday, solar energy strikes the earth with...Ch. 9 - Prob. 40EAPCh. 9 - Prob. 41EAPCh. 9 - Prob. 42EAPCh. 9 - 43. A 1000 kg elevator accelerates upward at 1.0...Ch. 9 - 44. a. Starting from rest, a crate of mass m is...Ch. 9 - Prob. 45EAPCh. 9 - 46. A particle of mass m moving along the x-axis...Ch. 9 -
47. A ball shot straight up with kinetic energy...Ch. 9 - 48. A pile driver lifts a 250 kg weight and then...Ch. 9 - Prob. 49EAPCh. 9 -
50. You’re fishing from a tall pier and have...Ch. 9 - Hook’s law describes an ideal spring. Many real...Ch. 9 -
52. The force acting on a particle is Fx =...Ch. 9 - 53. The gravitational attraction between two...Ch. 9 -
54. An electric dipole consists of two equal...Ch. 9 - Prob. 55EAPCh. 9 -
56. When a 65 kg cheerleader stands on a...Ch. 9 - Prob. 57EAPCh. 9 - Prob. 58EAPCh. 9 -
59. A horizontal spring with spring constant 250...Ch. 9 - 60. A 90 kg firefighter needs to climb the stairs...Ch. 9 - Prob. 61EAPCh. 9 - 62. When you ride a bicycle at constant speed,...Ch. 9 -
63. A farmer uses a tractor to pull a 150 kg...Ch. 9 - Prob. 64EAPCh. 9 - Prob. 65EAPCh. 9 - Prob. 66EAPCh. 9 - In problems 67 through 69 you are given the...Ch. 9 - Prob. 68EAPCh. 9 - Prob. 69EAPCh. 9 - Prob. 70EAPCh. 9 - Prob. 71EAPCh. 9 - Prob. 72EAP
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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 particle is subject to a force Fx that varies with position as shown in Figure P7.9. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?arrow_forwardThe 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.arrow_forward
- As 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_forwardSuppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0° slope at constant speed, as shown in Figure 7.37. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction 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?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
- The 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_forwardGive an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work.arrow_forwardA sled of mass 70 kg starts from rest and slides down a 10 incline 80 m long. It then travels for 20 m horizontally before starting back up an 8° incline. It travels 80 m along this incline before coming to rest. What is the magnitude of the net work done on the sled by friction?arrow_forward
- a shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed, (a) Find the work done by the shopper as she moves down a 50.0-m length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesnt change, would the shoppers applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?arrow_forwardA 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_forwardA student expends 7.5 W of power in lifting a textbook 0.50 m in 1.0 s with a constant velocity. (a) How much work is done, and (b) how much does the book weigh (in newtons)? The answers to Confidence Exercises may be found at the back of the book.arrow_forward
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