You have a summer job at your university’s
Want to see the full answer?
Check out a sample textbook solutionChapter 7 Solutions
Essential University Physics: Volume 1 (3rd Edition)
Additional Science Textbook Solutions
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
University Physics Volume 2
University Physics Volume 1
Tutorials in Introductory Physics
College Physics (10th Edition)
An Introduction to Thermal Physics
- Figure P9.65A shows a crate attached to a rope that is extended over an ideal pulley. Boris pulls on the other end of the rope with a constant force until the crate has risen a total distance of 6.53 m (Fig. P9.65B). If the crate has a mass of 81.36 kg, what is the average power exerted by Boris, assuming he accomplishes the task in 5.33 s? FIGURE P9.65arrow_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_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_forward
- The motion of a box of mass m = 2.00 kg along the x axis can be described by the function x = 4.00 + 3.00t2+ 2.00t3, where x is in meters and t is in seconds. a. What is the kinetic energy of the box as a function of time? b. What are the acceleration of the box and the force acting on the box as a function of time? c. What is the power delivered to the box as a function of time? d. What is the work performed on the particle during the time interval t = 1.00 s to t = 3.00 s?arrow_forward(a) What is the power output in watts and horsepower of a 70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00 s? (b) Considering the amount of power generated, do you think a well-trained athlete could do this repetitively for long periods of time?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 large cruise ship of mass 6.50 107 kg has a speed of 12.0 m/s at some instant. (a) What is the ships kinetic energy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of 2.50 km?arrow_forwardEstimate the power required for a boxer to jump rope.arrow_forwardBoxes A and B as shown have masses of 12.0 kg and 4.0 kg, respectively. The two boxes are released from rest. Use conservation of energy to find the boxes’ speed when box B has fallen a distance of 0.50 m. Assume a frictionless upper surface.arrow_forward
- A 2.40×10²N force is pulling an 85.0kg refrigerator across a horizontal surface. The force acts on an angle that is 20.0° above the horizontal. The coefficient of kinetic friction is 0.200. After the refrigerator moves for 8.00 meters, find the work done by the pulling force. What is the work done by the kinetic friction? If the refrigerator is initially at rest, use the work-kinetic energy model to find the final speed of the refrigerator. Answer: the work done by the pulling force will be 1.80×10°J. The work done by the kinetic friction will be 1.20×10'J. The net work done will be 6.01×10²J. As a result, the final speed will be v=3.76ms.arrow_forwardA brick lies perilously close to the edge of the flat roof of a building. The roof edge is 50 ft above street level, and the brick has 390.0 J of potential energy with respect to street level. Someone edges the brick off the roof, and it begins to fall. What is the brick’s kinetic energy when it is 35 ft above street level? What is the brick’s kinetic energy the instant before it hits the street surface?arrow_forwardA 20 N object is lifted a distance of 5.0 meters. The object has 100.0 N of potential energy. True or falsearrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning