The figure here shows a plot of potential energy U versus position x of a 0.891
ID: 1630595 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.891 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and UC = 45.0 J. The particle is released at x = 4.50 m with an initial speed of 7.51 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 7.51 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?
Explanation / Answer
Given that,
m = 0.891 kg
u = 7.51 m/s
initial KE is,
KEi = (1/2)mu^2
KEi = (1/2)*0.891*(7.51)^2 = 25.126 J
PE at x = 4.5 m,
PEi = 15 J
so, initial total energy Ei = 40.126 J
(a)
at x = 1 m, PE = 35 J
From energy cnservation, KE at x = 1,
KEf = 40.126 - 35 = 5.126 J
5.126 = (1/2)*0.891*v^2
so, speed at this point = v = 3.392 m/s
(b)
force on the particle as it begins to move to the left of x = 4.00 m,
F = -dU / dx
F = (35 - 15) / (2-4)
F = 10 N
(c)
direction of force - along +x axis
(d)
When particle reach x = 7 m,
PE = 45 J
As we can that PE > TE of particle . so will not reach at the this point.
Turning point will be b/w 5 m and 6 m.
Let, turning point = x
TEi = UA + (UC - UA)*(x - 5)
40.126 = 15 + 30*(x - 5)
x = 5.83 m
(e)
force on the particle as it begins to move to the right of x = 5.00 m,
F = -dU / dx = (45 - 15) / (6-5)
F = 30 N
(f)
direction of force = along -x axis.