A particle of mass m = 10 kg moves along the x-axis. A graph of its potential energy as a function of x is shown. (a) If the particle is released at rest at x = 3.0 m, what direction will it move (if at all)? (b) If the particle is released at x = 1.0 m with velocity v = -2.0 m/s, what speed will it have at x=0? (c) In the case described in part (b), what are the turning points, if any, of the motion?

A particle of mass 18.8 kg moves along the x-axis with a potential energy U whose dependence on position x is shown in the figure. If the total mechanical energy E of the particle is 110 J, what is the speed in m/s of the particle at 7 m? 15 U (J) 10 V. 5 4 8 10 X (m) 00 6. 20

The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C,where U will be in joules when x is in meters. A, B, and C are constants.  What is the force on this particle, in newtons, at x+39cm, if the constants are A= 2.2 J/m^2 , B= 2.2 J/m, and C=6.5 J?

The figure below shows a plot of potential energy U versus position x of a 0.76 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are UA = 20.0 J, UB = 35.0 J, and Uc = 45.0 J. Uc UB UA 4 х (m) The particle is released at x = 4.5 m with an initial speed of 6.0 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? m (b) 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? magnitude direction +X Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 6.0 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? m (d) 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? magnitude…