Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 8, Problem 15P
You have spent a long day skiing and are tired. You are standing at the top of a hill, looking at the lodge at the bottom of the hill. You are so tired that you want to simply start from rest and coast down the slope, without pushing with your poles or doing anything else to change your motion. You want to let gravity do all the work! You have a choice of two trails to reach the lodge. Both trails have the same coefficient of friction μk. In addition, both trails represent the same horizontal separation between the initial and final points. Trail A has a short, steep downslope and then a long, flat coast to the lodge. Trail B has a long, gentle downslope and then a short remaining flat coast to the lodge. Which trail will result in your arriving at the lodge with the highest final speed?
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Chapter 8 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 8.1 - Consider a block sliding over a horizontal surface...Ch. 8.2 - A rock of mass m is dropped to the ground from a...Ch. 8.2 - Three identical balls are thrown from the top of a...Ch. 8.3 - You are traveling along a freeway at 65 mi/h. Your...Ch. 8 - Prob. 1PCh. 8 - A 20.0-kg cannonball is fired from a cannon with...Ch. 8 - A block of mass m = 5.00 kg is released from point...Ch. 8 - At 11:00 a.m, on September 7, 2001, more than one...Ch. 8 - A light, rigid rod is 77.0 cm long. Its top end is...Ch. 8 - Prob. 6P
Ch. 8 - A crate of mass 10.0 kg is pulled up a rough...Ch. 8 - A 40.0-kg box initially at rest is pushed 5.00 m...Ch. 8 - Prob. 9PCh. 8 - As shown in Figure P8.10, a green bead of mass 25...Ch. 8 - At time ti, the kinetic energy of a particle is...Ch. 8 - A 1.50-kg object is held 1.20 m above a relaxed...Ch. 8 - Prob. 13PCh. 8 - An 80.0-kg skydiver jumps out of a balloon at an...Ch. 8 - You have spent a long day skiing and are tired....Ch. 8 - The electric motor of a model train accelerates...Ch. 8 - An energy-efficient lightbulb, taking in 28.0 W of...Ch. 8 - An older-model car accelerates from 0 to speed v...Ch. 8 - Prob. 19PCh. 8 - There is a 5K event coming up in your town. While...Ch. 8 - Prob. 21PCh. 8 - Energy is conventionally measured in Calories as...Ch. 8 - A block of mass m = 200 g is released from rest at...Ch. 8 - Prob. 24APCh. 8 - Prob. 25APCh. 8 - Review. As shown in Figure P8.26, a light string...Ch. 8 - Consider the blockspringsurface system in part (B)...Ch. 8 - Why is the following situation impossible? A...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - As the driver steps on the gas pedal, a car of...Ch. 8 - As it plows a parking lot, a snowplow pushes an...Ch. 8 - Prob. 33APCh. 8 - Prob. 34APCh. 8 - A horizontal spring attached to a wall has a force...Ch. 8 - Prob. 36APCh. 8 - Prob. 37APCh. 8 - Review. Why is the following situation impossible?...Ch. 8 - Prob. 39APCh. 8 - A pendulum, comprising a light string of length L...Ch. 8 - Prob. 41APCh. 8 - Prob. 42APCh. 8 - Prob. 43APCh. 8 - Starting from rest, a 64.0-kg person bungee jumps...Ch. 8 - Prob. 45CPCh. 8 - A uniform chain of length 8.00 m initially lies...Ch. 8 - Prob. 47CP
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- 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_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 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 P5.8. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity, and (d) the net force on the block. Figure P5.8arrow_forward
- A 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 ail angle = 25 below the horizontal as shown in Figure P7.5. 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.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_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_forward
- Consider 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_forwardThe system shown in Figure P5.43 is used to lift an object of mass m = 76.0 kg. A constant downward force of magnitude F is applied to the loose end of the rope such that the hanging object moves upward at constant speed. Neglecting the masses of the rope and pulleys, find (a) the required value of F, (b) the tensions T1, T2, and T3, and (c) the work done by the applied force in raising the object a distance of 1.80 m. Figure P5.43arrow_forwardA boy starts at rest and slides down a frictionless slide as in Figure P5.64. The bottom of the track is a height h above the ground. The boy then leaves the track horizontally, striking the ground a distance d as shown. Using energy methods, determine the initial height H of the boy in terms of h and d. Figure P5.64arrow_forward
- Repeat 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_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_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_forward
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