Chapter 08, Problem 039 The figure here shows a plot of potential energy U versu

Question

Chapter 08, Problem 039 The figure here shows a plot of potential energy U versus position x of a 0.852 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15.0 J UB 35.0 J and UC 45.0 J. The particle is released at x 4.50 m with an initial speed of 7 28 m s headed in the negat exdirec on. (a) If the particle can reach x - 1.00 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 = 4.00 m? Suppose, instead, the particle is headed in the positivex direction when ti released at r 4,5 m at speed 7.28 m/s. d If the particle can reach x = 7.00 m, what is its speed there, and if it cannot what is its turning point at are the e magnitude and direction of the force on the particle as it begins to move to the right of x 5.00 m? [--+ 6 x (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 + ½*0.852kg*(7.28m/s)2

                           = 37.6 J

(a) at x = 1 m, U = 35 J < TME, so the particle can reach there.
Its remaining KE = 2.6 J = ½ * 0.852kg * v²
v = 2.46 m/s

Let's find the acceleration from x = 4m to x = 2 m:
v² = u² + 2as
(2.46 m/s)² = (7.28m/s)² + 2 * a * 2m
a = -11.73 m/s² (against the direction of motion, or rightward).

(b) avg F = ma

               = 0.852kg * 11.73m/s²

                = 10 N (in magnitude)

(c) and 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 = 37.6 J, which occurs near
x = 5m + 1m*(37.6 - 15) / (45-15)

   = 5.75 m

(d) v² = u² + 2as
(0 m/s)² = (7.28m/s)² + 2 * a * 1m
a = -26.5 m/s² (against the direction of motion, leftward).

(e) F = 0.852kg * 26.5 m/s² = 22.6 N in magnitude
(f) and direction "to the left"

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