Chapter 08, Problem 038 The figure shows a plot of potential energy U versus pos

Question

Chapter 08, Problem 038 The figure shows a plot of potential energy U versus position x of a 0.270 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 6.50 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side? Please explain equations

Explanation / Answer

here,

UB = 12 J and KEB = 6.5 J

m = 0.27 kg

(a)

at x = 3.5 ,

let the speed be v

using conservation of energy

UB + KEB = UA + 0.5 * m * v^2

12 + 6.5 = 9 + 0.5 * 0.27 * v^2

v = 8.39 m/s

(b)

at x = 6.5 ,

let the speed be v

using conservation of energy

UB + KEB = U0 + 0.5 * m * v^2

12 + 6.5 = 0 + 0.5 * 0.27 * v^2

v = 11.71 m/s

(c)

when the rate of change of potential energy changes ,

that point is known as the turning point

the turning point on the right side is x = 7 m

(d)

the turning point on the left side is at x = 4 m

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