The figure shows a plot of potential energy U versus position x of a 0.220 kg pa

Question

The figure shows a plot of potential energy U versus position x of a 0.220 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?

Explanation / Answer

The figure is not provided. It requires a login to wiley plus. Please upload the image without which the answer cannot be solved.

However giving you the concept of solving this problem

You need to apply the conservation of energy.

The total energy at From the point where it is released it 12 + 6.5 = 18.5 J Which will remain constant.

KE = 1/2*m*v^2

and PE = mgh

So you can determine the Kinetic Energy at points A, C and D.

You can calculate the velocity also using the equation of Kinetic Energy.

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