The figure here shows a plot of potential energy U versus position x of a 0.858
ID: 1339453 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.858 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 7.1 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 7.1 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
Since non-conservative forces are not in play, the total mechanical energy (TME) of the system is constant.
Initially, this TME = U + KE = 15J + (1/2)*0.858kg*(7.1m/s)2 = 36.6259 J
(a) at x = 1 m, U = 35 J < TME, so the particle can reach there.
Its remaining K.E. =1.6259 J = 0.5* 0.858kg * v2
v = 1.9468 m/s
Let's find the acceleration from x = 4m to x = 2 m:
v2 = u2 + 2as
(1.9468m/s)2= (7.1m/s)2 + (2 * a * 2m )
a = -11.6550 m/s2(against the direction of motion, or rightward).
(b) avg F = ma = 0.858kg * 11.6550m/s² = 9.9999 N (in magnitude)
(c) the force obtained in (b) has a direction to the right, against the direction of motion.
(d) The particle cannot reach 7 m, or even 6 m, where U = 45 J > TME.
It can reach only U = 36.6259 J, which occurs near
x = 5m + 1m*(36.6259 -15)/(45-15) = 5.7209 m
(e) v2 = u2+ 2as
(0 m/s)² = (7.1m/s)² + 2 * a * 1m
a = -25.2050 m/s² (against the direction of motion, leftward).
F = 0.858kg * 25.2050m/s² = 21.6259N in magnitude
(f) the direction of the force obtained in (e) is "to the left"