The figure below shows a plot of potential energy U versus position x of a 1. 46

Question

The figure below shows a plot of potential energy U versus position x of a 1. 46 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 = 40 J. The particle is released at x = 4. 5 m with an initial speed of 6. 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 6. 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

intitial kinetic energy=0.5*1.46*6.5^2=30.8425 J initial potential energy=U1=5 J so net energy=E=35.8425 J a) if it reaches x=1m, its potential energy =U2=30 J so it has 5.8425 J kinetic energy left with it. so it CAN reach c=1 m speed=v(let) then 0.5*1.46*v^2=5.8425 v=2.83 m/s b)force=-du/dx so to the left, du/dx=(U1-U2)/(4-2)=-12.5 so force=12.5 N direction is along +ve x-axis c) if it reaches x=7, it has potential eenrgy=40 J but total energy=35.8425 J as kinetic energy cant be negative(simply speed cant be -ve) it will stop some point in between 5 m and 6 m now slope of potential energy curve between 5 m and 6 m=(U3-U1)/(6-5)=35 J/m so it will stop at x=((35.8425-5)/(35))+5=5.8812 m d) to the right of x=5 m, force=-du/dx =-35 N so force will be along -ve x-direction.

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