need help in part d only. The figure here shows a plot of potential energy U ver
ID: 1641945 • Letter: N
Question
need help in part d only.
The figure here shows a plot of potential energy U versus position x of a 0.885 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are U_A = 15.0 J, U_B = 35.0 J and U_C = 45.0. The particle is released at x = 4.50 m with an initial speed of 7.50 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.50 m/s. (d) If the particle can reach x = 7.50 m/s, particle can reach x = 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.00 m? (a) Number Unit (b) Number Unit (c) (d) Number UnitExplanation / Answer
Since only conservative forces are involved, the mechanical energy of the particle is conserved.
Total energy of the particle at x = 4.50 m is,
EA = UA + mvA2/2 = 15.0 + (0.885 * 7.502 / 2) = 39.9 J
Since the potential energy of the particle at x = 7.00 m is 45.0 J > 39.9 J, the particle cannot reach there.
Potential energy between x = 5.00 m and x = 6.00 m is given by,
U = 15 + [(45 - 15) / (6 - 5)](x - 5)
=> U = 15 + 30(x - 5) = 30x - 135
At the turning point,
30x - 135 = 39.9
=> x = 5.83 m
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