The figure here shows a plot of potential energy U versus position x of a 0.851
ID: 1643763 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.851 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 J. The particle is released at x = 4.50 m with an initial speed of 7.49 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 7.49 m/s. (d) If the 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 m/s (b) Number Unit (c) [positive (+x) (d) Number Unit (e) Number Unit (f) negative (-x) Click if you would like to Show Work for this question: Open Show WorkExplanation / Answer
(a) Inital mechanical energy of the particle,
Ei = UA + mvi2/2 = 15.0 + (0.851 * 7.492 / 2) = 38.9 J
Minimum mechanical energy at x = 1 m is 35.0 J, so the particle can reach there.
Using conservation of energy,
Mechancial energy at 1.00 m = Mechanical energy at 4.50 m
UB + mv2/2 = 38.9
=> 35.0 + (0.851 * v2 / 2) = 38.9
=> v = [2 * (38.9 - 35.0) / 0.851]1/2 = 3.03 m/s
(b) F = -dU/dx
So, the force acting on the particle is the negative of the slope of the U-x graph.
For x < 4.00 m,
F = -(35.0 - 15.0) / (2 - 4) = 10.0 N
(c) This force is acting to the right since the sign is positive.
(d) Minimum mechanical energy at x = 7.00 m is 45.0 J, which is greater than particle's initial energy i.e. 38.9 J
Hence, the particle can't reach x = 7.00 m.
The turning point of the particle will be where U = 38.9 J. So, equating slopes,
(38.9 - 15) / (x - 5) = (45 - 15) / (6 - 5)
=> 23.9 / (x - 5) = 30
=> x = 5 + 23.9/30 = 5.80 m
(e) To the right of x = 5 m,
F = -dU/dx = -(45 - 15) / (6 - 5) = -30 N
So, magnitude of force = 30 N
(f) Direction is to the left, since the sign is negative.