A particle moves along the x-axis while acted on by a single conservative force parallel to the x-axis. The force corresponds to the potential-energy function graphed in ?Fig. P7.76?. The particle is released from rest at point A. (a) What is the direction of the force on the particle when it is at point A? (b) At point B? (c) At what value of x is the kinetic energy of the particle a maximum? (d) What is the force on the particle when it is at point C? (e) What is the largest value of x reached by the particle during its motion? (f) What value or values of x correspond to points of stable equilibrium? (g) Of unstable equilibrium?

Solution 86P Step 1: Etotal K + U Initially,the particle is released from the rest position point A means K = 1/2 mv = 0 Step 2: (a).F = dU/dX at A. slope is negative so F is positive or repulsive,so A to right (b). at B. slope is positive so F is negative or attractive,Bo F to left Force always push particle to lowest potential,increasing kinetic energy. Step 3: (c). K = E U Kmax at 0.75 at the minimum of U(x) K at x = A min K = 0 since E total U( x = A) Step 4: (d).The force on the particle at point C is At point C dU/dX = 0 then F = 0 F = 0 means unstable equilibrium.coin on edge is stable.