The figure here shows a plot of potential energy U versus position x of a 0.887
ID: 2253205 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.887 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 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. (a) 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 (b) magnitude and (c) 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. (d) 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 (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.0 m?
Explanation / Answer
a) Ei = 15 + 0.5*0.887*7.5^2=39.94 J
since Ei>Ub it can reach
Ef = 1/2*0.887*v^2 + 35 = 39.94
v=3.34 m/s
b) F = dU/dx = (35-15)/2 = 10 N
c) to the right
d) since Ei<Uc it can't reach there
39.94 = 15 + (45-15)/1*x
x=0.831
so turning point is 5.831 m
e) F = dU/dx = (45-15)/1 = 30 N
f) to the left