The figure below shows a plot of potential energy U versus position x of a 1.50

Question

The figure below shows a plot of potential energy U versus position x of a 1.50 kg particle that can travel only along an x axis. (Neoconservative forces are not involved.) In the graphs, the potential energies are U_1 = 15 J, U_2 = 30 J, and U_3 = 40 J. The particle is released at x = 4.5 m with an initial speed of 7.5 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? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.5 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?

Explanation / Answer

(a) as non conservative forces are not involved herefore energy is conserved

kinetic energy at x = 4.5

= 1/2 * mass *v2 = 1/2 * 1.5 * 7.52 = 42.19 J

potential energy = 15 J [ from the graph]

let velocity at x = 1.0 m be v0

(mechanical energy at x= 4.5 m) = (mechanical energy at x = 1m) [ law of conservation of energy]

15+ 42.19 =30 + 1/2 m * v02

v = 6.02 m/s

(b) initial velocity = 7.5 m/s

final velocity = 6.02 m/s

acceleration = (6.02 - 7.5) / (4-2) = -0.74 m /s2

force = mass x acceleration = 1.5 x (-0.74) = -0.49 N

negative sign shows force is towards right

(c)total potential energy at x = 7m is 40 J

let the final velocity be v0

(mechanical energy at x= 4.5 m) = (mechanical energy at x = 7m) [ law of conservation of energy]

15+ 42.19 =40 + 1/2 m * v02

v0 = 4.79 m/s

(d)

initial velocity = 7.5 m/s

final velocity = 4.79 m/s

acceleration = (4.79 - 7.5) / (4-2) = -1.36 m /s2

force = mass x acceleration = 1.5 x (-1.36) = - 2.03N

negative sign shows force is towards left

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