The figure below shows a plot of potential energy U versus position x of a 1.08

Question

The figure below shows a plot of potential energy U versus position x of a 1.08 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 5 J, U2 = 30 J, and U3 = 50 J.

The particle is released atx=4.5m with an initial speed of6.0m/s, headed in the negativexdirection.

(a) If the particle can reachx= 1.0 m, what is its speed there, and if it cannot, what is its turning point?

(b) What are the magnitude and direction of the force on the particle as it begins to move to the left ofx= 4.0 m?

Suppose, instead, the particle is headed in the positivexdirection when it is released atx=4.5m at speed6.0m/s.

(c) If the particle can reachx= 7.0 m, what is its speed there, and if it cannot, what is its turning point?

(d) What are the magnitude and direction of the force on the particle as it begins to move to the right ofx= 5.0 m?

Explanation / Answer

since non conservative forces are not present, hence energy will remain conserved. hence , when particle reaches from x = 4.5 to x = 1.0, its potential energy is increased and kinetic energy will decrease decrease in KE = increase in PE = U2 - U1 = 25 J 0.5*1.08 *( 6^2 - v_f^2) =25 6^2 - v_f^2 = 46.29 V_F^2 = -10.29 (NOT POSSIBLE) hence particle will not be able to reach , it will turn somewhere b/w x=2 to x =4 at the point of turning , we can write 0.5*1.08*6^2 = U - U1 U = 24.44 J (energy at the turning point ) now the location of the turning point can be found from the slop of the graph b/w x=2 to x =4, (24.44- 5)/(x - 4.5) = (30-5)/(2-4.5) x = 2.556 m b) force = -dU/dx = - (30-5)/(2-4.5) = 10 N (directed along +ve x- axis) C) decrease in KE = increase in PE = U3 - U1 = 45 J 0.5*1.08 *( 6^2 - v_f^2) =45 6^2 - v_f^2 = 66.29 v_f^2 is comint to be negative( not possible) hence particle will not be able to reach , it will turn somewhere b/w x=4.5 to x =6 at the point of turning , we can write 0.5*1.08*6^2 = U - U1 U = 24.44 J (energy at the turning point ) now the location of the turning point can be found from the slop of the graph b/w x=4.5 to x =6, (24.44- 5)/(x - 4.5) = (50-5)/(6-4.5) x =5.58 m d) force = -dU/dx = - 50-5)/(6-4.5) = -18N magnitude = 18 N direction along -ve x-axis

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