A 3.0 kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at time t = 0, a steady wind pushes on the lunchbox in the negative direction of the x axis. The figure shows the position x of the lunchbox as a function of time t as the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a) t = 1.0 s and (b) t = 5.0 s. (c) How much work does the force from the wind do on the lunchbox from t = 1.0 s to t = 5.0 s?

A 3.0 kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at time t = 0, a steady wind pushes on the lunchbox in the negative direction of the x axis. The figure shows the position x of the lunchbox as a function of time t as the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a) t = 1.0 s and (b) t = 5.0 s. (c) How much work does the force from the wind do on the lunchbox from t = 1.0 s to t = 5.0 s?

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter9: Energy In Nonisolated Systems

Section: Chapter Questions

Problem 85PQ: The motion of a box of mass m = 2.00 kg along the x axis can be described by the function x = 4.00 +...

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(a) A child slides down a water slide at an amusement park from an initial height h. The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is 12.0 m.
A puck of mass 0.170 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and puck of 0.150. If the puck is moving at an initial speed of 12.0 m/s, (a) what is the force of kinetic friction? (b) What is the acceleration of the puck? (c) How long does it take for the puck to come to rest? (d) What distance does the puck travel during that time? (e) What total work does friction do on the puck? (f) What average power does friction generate in the puck during that time? (g) What instantaneous power does friction generate in the puck when the velocity is 6.00 m/s? (See Sections 2.5, 4.6, 5.1, and 5.6.)
(a) A child slides down a water slide at an amusement park from an initial height h. The slide can be considered frictionless because of the water flowing down it. Can the equation for conservation of mechanical energy be used on the child? (b) Is the mass of the child a factor in determining his speed at the bottom of the slide? (c) The child drops straight down rather than following the curved ramp of the slide. In which case will he be traveling faster at ground level? (d) If friction is present, how would the conservation-of-energy equation be modified? (e) Find the maximum speed of the child when the slide is frictionless if the initial height of the slide is 12.0 m.
Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch?
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?
A child of mass m starts from rest and slides without friction from a height h along a slide next to a pool (Fig. P7.27). She is launched from a height h/5 into the air over the pool. We wish to find the maximum height she reaches above the water in her projectile motion. (a) Is the childEarth system isolated or nonisolated? Why? (b) Is there a nonconservative force acting within the system? (c) Define the configuration of the system when the child is at the water level as having zero gravitational potential energy. Express the total energy of the system when the child is at the top of the waterslide. (d) Express the total energy of the system when the child is at the launching point. (e) Express the total energy of the system when the child is at the highest point in her projectile motion. (f) From parts (c) and (d), determine her initial speed vi at the launch point in terms of g and h. (g) From parts (d), (e), and (f), determine her maximum airborne height ymax in terms of h and the launch angle . (h) Would your answers be the same if the waterslide were not frictionless? Explain. Figure P7.27
A 60.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.0 m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athletes speed is 9.0 m/s as y = 0. What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (e) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.
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.
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 P9.30ANS