he figure below shows a plot of potential energy U versus position x of a 1.14 k
ID: 1287570 • Letter: H
Question
he figure below shows a plot of potential energy U versus position x of a 1.14 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, the potential energies are U1 = 5 J, U2 = 20 J, and U3 = 35 J.
The particle is released at x = 4.5 m with an initial speed of 6.0 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?
Answer:______(m or m/s)
(b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m?
Answer:_____ N in the _______(positive or negative) direction.
Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 6.0 m/s.
(c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point?
Answer:______(m or m/s)
(d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m?
Answer:_____ N in the _______(positive or negative) direction.
Explanation / Answer
a) initial kinetic energy = mv^2 /2 = 1.14 x 6^2 /2 =20.52 J
initial total mechanical energy = P.E. + K.E
= 20.52 + 5 = 25.52 J
at x=1 m
P.E> = 20 J
K.E. = 25.52 - 20 = 5.52 = 1.14v^2 /2
v = 3.11 m/s
b) F = - dU /dx
so F = - slope = - 1 x (20-5) / (2-4) = 7.5 N
c) P.E. at x= 7 , is 35J but total M.E. is only 25.52 J .
so it cant reach at 7.
it can have max. P>E. = 25.52 J
so, (U - 5) = 30(x - 5)
so , (25.52 - 5) = 30(x -5)
x = 5.68 m
d) F = - dU/dx
F = - slope = - 30 N
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