The figure shows a plot of potential energy U versus position x of a 0.300 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.300 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9.00 J, UC = 20.0 J and UD = 24.0 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12.0 J, with kinetic energy 5.00 J. What is the speed of the particle at (a)x = 3.50 m and (b)x = 6.50 m? What is the position of the turning point on (c) the right side and (d) the left side?

I have answer a=7.30. b=11.5, c=7.83, d=1.36. b-c-d are incorrect and cannot figure out why. my answer needs to be 3 significant figures.

Explanation / Answer

A)

Kinetic energy, E = 12 + 5 - 9 = 8 J

Kinetic energy, E = 0.5 mv^2

Velocity, v = sqrt(2E/m) = sqrt (2 x 8/0.3) = 7.30 m/s

B)

From law of conservation of energy  

12 + 5 = 0 + E

E = 17 J

Velocity, v = sqrt(2 x 17/0.3) = 10.65 m/s

C)

17/(x - 7) = (24 - 17)/(8 - x)

X = 7.708 m

D)

(17 - 20)/(x - 1) = (9 - 17)/(3 - x)

X = 1.545 m

Comment in case any doubt please rate my answer....

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