The figure below shows a plot of potential energy U versus position x of a 0.98

Question

The figure below shows a plot of potential energy U versus position x of a 0.98 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 10J, U2 = 25J, and U3 = 45J. The particle is released at x = 4.5 m with an initial speed of 9.5 m/s, headed in the negative x direction. If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? N in the direction. Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 9.5 m/s. If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? N in the direction.

Explanation / Answer

The kinetic energy at x = 4.5m is    K = (1/2)(0.98kg)(9.5m/s)^2        = 44.22kg Potential energy is U = 10J total energy is E = 44.22 + 10 = 54.22 J at x =1, the potential energy is 25J so kinetic energyis         K = 54.22J - 25 =29.22 J speed of the particle is     v = 2K/ m = 2(29.22) /0.98        = 7.72m/s -------------------------------------------------------------------------------------------------------------- b) The force acting on the particle is        F = -dU/ dx           = -(25-10) / 2-4          = 7.5 N The force is positive , so it is positive x direction -------------------------------------------------------------------------------------------------------------------- c) The potential energy is less than the totalenergy         KE = 54.2223 -45              = 9.2223 J    The speed of the particle is      v = 2k/m         = 2(9.2223)/0.98)        = 4.3383m/s ------------------------------------------------------------------------------------------------------------------- b) Force is         F = - (45 -10) /(6-5)            =- 35 N    Force is 30N moving on negative xdirection

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