Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
4th Edition
ISBN: 9780134110684
Author: Randall D. Knight (Professor Emeritus)
Publisher: PEARSON
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Chapter 10, Problem 31EAP
A system in which only one particle can move has the potential energy shown in FIGURE EX10.31. What is the y-component of the force on the particle at y = 0.5 m and 4 m?
FIGURE EX10.31
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Chapter 10 Solutions
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
Ch. 10 - Prob. 1CQCh. 10 - Can kinetic energy ever be negative? Can...Ch. 10 - Prob. 3CQCh. 10 - 4. The three balls in FIGURE Q1O.4, which have...Ch. 10 - Rank in order, from most to least, the elastic...Ch. 10 - 6. A spring is compressed 1.0 cm. How far must you...Ch. 10 - Prob. 7CQCh. 10 - A particle with the potential energy shown in...Ch. 10 - A compressed spring launches a block up an...Ch. 10 - 10. A process occurs in which a system’s potential...
Ch. 10 - A process occurs in which a system’s potential...Ch. 10 - FIGURE Q10.12 is the energy bar chart for a...Ch. 10 - Prob. 13CQCh. 10 - Object A is stationary while objects B and C are...Ch. 10 - Prob. 2EAPCh. 10 - 3. The lowest point in Death Valley is 85 m below...Ch. 10 - Prob. 4EAPCh. 10 - Prob. 5EAPCh. 10 - 6. What height does a frictionless playground...Ch. 10 - 7. A 55 kg skateboarder wants to just make it to...Ch. 10 - Prob. 8EAPCh. 10 - A pendulum is made by tying a 500 g ball to a...Ch. 10 - A 20 kg child is on a swing that hangs from...Ch. 10 - A 1500 kg car traveling at 10 m/s suddenly runs...Ch. 10 - Prob. 12EAPCh. 10 - A cannon tilted up at a 30° angle fires a cannon...Ch. 10 - In a hydroelectric dam, water falls 25 m and then...Ch. 10 - How far must you stretch a spring with k = 000 N/m...Ch. 10 - A stretched spring stores 2.0 J of energy. How...Ch. 10 - A student places her 500 g physics book on a...Ch. 10 - A block sliding along a horizontal frictionless...Ch. 10 - A 10 kg runaway grocery cart runs into a spring...Ch. 10 - As a 15,000 kg jet plane lands on an aircraft...Ch. 10 - The elastic energy stored in your tendons can...Ch. 10 - The spring in FIGURE EX10.22a is compressed by ?x....Ch. 10 - The spring in FIGURE EXIO.23a is compressed by ?x....Ch. 10 - FIGURE EX10.24 is the potential-energy diagram for...Ch. 10 - Prob. 25EAPCh. 10 - In FIGURE EX10.26, what is the maximum speed of a...Ch. 10 - Prob. 27EAPCh. 10 - FIGURE EX10.28 shows the potential energy of a 500...Ch. 10 - In FIGURE EX10.28, what is the maximum speed a 200...Ch. 10 - A system in which only one particle can move has...Ch. 10 - A system in which only one particle can move has...Ch. 10 - A particle moving along the y-axis is in a system...Ch. 10 - A particle moving along the x-axis is in a system...Ch. 10 - FIGURE EX10.34 shows the potential energy of a...Ch. 10 - A particle moves from A to D in FIGURE EX10.35...Ch. 10 - A force does work on a 50 g particle as the...Ch. 10 - A system loses 400 J of potential energy. In the...Ch. 10 - What is the final kinetic energy of the system for...Ch. 10 - How much work is done by the environment in the...Ch. 10 - A cable with 20.0 N tension pulls straight up on a...Ch. 10 - A very slippery ice cube slides in a vertical...Ch. 10 - A 50 g ice cube can slide up and down a...Ch. 10 - You have been hired to design a spring-launched...Ch. 10 - It’s been a great day of new, frictionless snow....Ch. 10 - Prob. 45EAPCh. 10 - A 1000 kg safe is 2.0 m above a heavy-duty spring...Ch. 10 - You have a ball of unknown mass, a spring with...Ch. 10 - Sam, whose mass is 75 kg, straps on his skis and...Ch. 10 - A horizontal spring with spring constant 100 N/m...Ch. 10 - Truck brakes can fail if they get too hot. In some...Ch. 10 - Prob. 51EAPCh. 10 - Use work and energy to find an expression for the...Ch. 10 - Prob. 53EAPCh. 10 - The spring shown in FIGURE 10.54 is compressed 50...Ch. 10 - Prob. 55EAPCh. 10 - Prob. 56EAPCh. 10 - A system has potential energy U(x) = x + sin ((2...Ch. 10 - Prob. 58EAPCh. 10 - Prob. 59EAPCh. 10 - Prob. 60EAPCh. 10 - The potential energy for a particle that can move...Ch. 10 - A particle that can move along the x-axis...Ch. 10 - An object moving in the xy-plane is subjected to...Ch. 10 - An object moving in the xy-plane is subjected to...Ch. 10 - Prob. 65EAPCh. 10 - In Problems 66 through 68 you are given the...Ch. 10 - Prob. 67EAPCh. 10 - Prob. 68EAPCh. 10 - A pendulum is formed from a small ball of mass m...Ch. 10 - Prob. 70EAPCh. 10 - Prob. 71EAPCh. 10 - Prob. 72EAPCh. 10 - The spring in FIGURE CP10.73 has a spring constant...Ch. 10 - A sled starts from rest at the top of the...
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- A 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_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_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
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- 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_forwardReview. A bead slides without friction around a loop-the-loop (Fig. P7.3). The bead is released from rest at a height h = 3.50R. (a) What is its speed at point ? (b) How large is the normal force on the bead at point if its mass is 5.00 g? Figure P7.3arrow_forwardA 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_forward
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Kinetic Energy and Potential Energy; Author: Professor Dave explains;https://www.youtube.com/watch?v=g7u6pIfUVy4;License: Standard YouTube License, CC-BY