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 3P
A block of mass m = 5.00 kg is released from point Ⓐ and slides on the frictionless track shown in Figure P8.3. Determine (a) the block’s speed at points Ⓑ and Ⓒ and (b) the net work done by the gravitational force on the block as it moves from point Ⓐ to point Ⓒ.
Figure P8.3
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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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- 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 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 is placed on top of a vertical spring, and the spring compresses. Figure P8.24 depicts a moment in time when the spring is compressed by an amount h. a. To calculate the change in the gravitational and elastic potential energies, what must be included in the system? b. Find an expression for the change in the systems potential energy in terms of the parameters shown in Figure P8.24. c. If m = 0.865 kg and k = 125 N/m, find the change in the systems potential energy when the blocks displacement is h = 0.0650 m, relative to its initial position. FIGURE P8.24arrow_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
- 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 jack-in-the-box is actually a system that consists of an object attached to the top of a vertical spring (Fig. P8.50). a. Sketch the energy graph for the potential energy and the total energy of the springobject system as a function of compression distance x from x = xmax to x = 0, where xmax is the maximum amount of compression of the spring. Ignore the change in gravitational potential energy. b. Sketch the kinetic energy of the system between these points the two distances in part (a)on the same graph (using a different color). FIGURE P8.50 Problems 50 and 79arrow_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
- 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 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_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_forwardThe Flybar high-tech pogo stick is advertised as being capable of launching jumpers up to 6 ft. The ad says that the minimum weight of a jumper is 120 lb and the maximum weight is 250 lb. It also says that the pogo stick uses a patented system of elastometric rubber springs that provides up to 1200 lbs of thrust, something common helical spring sticks simply cannot achieve (rubber has 10 times the energy storing capability of steel). a. Use Figure P8.32 to estimate the maximum compression of the pogo sticks spring. Include the uncertainty in your estimate. b. What is the effective spring constant of the elastometric rubber springs? Comment on the claim that rubber has 10 times the energy-storing capability of steel. c. Check the ads claim that the maximum height a jumper can achieve is 6 ft.arrow_forward
- 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_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
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