Student Workbook for Physics for Scientists and Engineers: A Strategic Approach, Vol 1. (Chs 1-21)
4th Edition
ISBN: 9780134110646
Author: Randall D. Knight (Professor Emeritus)
Publisher: PEARSON
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Chapter 10, Problem 32EAP
A particle moving along the y-axis is in a system with potential energy U = 4y3J, where y is in m. What is the y-component of the force on the particle at y = 0 m, 1 m, and 2 m?
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Chapter 10 Solutions
Student Workbook for Physics for Scientists and Engineers: A Strategic Approach, Vol 1. (Chs 1-21)
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 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_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_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_forward
- (a) Sketch a graph of the potential energy function U(x)=kx2/2+Aex2 where k , A, and are constants. (b) What is the force corresponding to this potential energy? (c) Suppose a particle of mass in moving with this potential energy has a velocity v when its position is x = . Show that the particle does not pass 2+2 through the origin unless Amv2=k22(1e a 2 ) .arrow_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_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
- 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_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_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_forward
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Mechanical work done (GCSE Physics); Author: Dr de Bruin's Classroom;https://www.youtube.com/watch?v=OapgRhYDMvw;License: Standard YouTube License, CC-BY