A particle of mass m = 1.0 kg moves along the x axis without friction through a
ID: 1324626 • Letter: A
Question
A particle of mass m = 1.0 kg moves along the x axis without friction through a region in which its potential energy U(x) varies as shown in the Figure. When the particle is at x = 7.0 m, its velocity is 2.550 m/s. What is the magnitude of the force that acts on the particle at this position?
B)
The particle will move back and forth in a limited region along the x axis. What is the smallest value of x which the particle ever has?
A particle of mass m = 1.0 kg moves along the x axis without friction through a region in which its potential energy U(x) varies as shown in the Figure. When the particle is at x = 7.0 m, its velocity is 2.550 m/s. What is the magnitude of the force that acts on the particle at this position? B) The particle will move back and forth in a limited region along the x axis. What is the smallest value of x which the particle ever has?Explanation / Answer
The Kinetic Energy at x = 7.0 is:
KE = (1/2)*m*v^2 = (1/2)*1*(2.550)^2 = 3.25J
This energy could be converted to additional potential energy as the particle makes a lower x excursion. The PE function has slope:
dU/dx = (-15 - (-5))/(10 - 2) = -10/8= -1.25
The PE at x=7 is:
PE(7) = -5 + (7 - 2)*(-1.25) = -11.25 J
If the available KE is converted to PE, the PE can go as low as:
PE( x_min) = -11.25 + 3.25 = -8 J
PE(x_min) = -8 = -5 + (x - 2)*(-1.25)
(x - 2)*(-1.25) = -3
(x - 2) = -3/(-1.25)
x = 2 + -3/(-1.25) = 4.4m ans
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