The figure here shows a plot of potential energy U versus position Times of a 0.
ID: 1369539 • Letter: T
Question
The figure here shows a plot of potential energy U versus position Times of a 0.858 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are U_A = 15 J, U_B =- 35 J and U_C = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.6 m/s, headed in the negative x direction, 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 magnitude and 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 7.6 m/s. 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 magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m?Explanation / Answer
a) The Kinetic energy at x = 4.5m = (1/2)mV2 = 0.5*0.858*7.62 = 24.779 J
The Potential Energy at this point = UA = 15 J.
So net mechanical energy at this point = KE + PE = 39.779 J
At x=1m the potential energy = UB = 35 J
Since there is only conservative forces working on it so the mechanical energy will be conserved.
so Kinetic energy at x = 1, KE = 4.779 J and speed = sqrt(4.779*2/0.858) = 3.338 m/s
b) The force is negative of the slope of U vs x curve.
so to the left of x= 4m the force will be 10 N
c) The force is in +x-direction.
d) The mechanical energy of the particle = 39.779 J
at x=7 , U is greater than the mechanical energy so it can't reach at x = 7m. Now to find the turning point at which the speed will be zero and so KE, the potential energy will be 39.779 J
U = 15 + 30 (x-5) = 39.779
so, x = 5+0.83 = 5.83m
e) when x is left to 5m the slope of U vs x curve is +30 so the force will be 30 N
f) As the slope is +ve, the direction of force will be negative x-direction.