A 7.53 kg particle is subjected to only conservative forces and its potential en

Question

A 7.53 kg particle is subjected to only conservative forces and its potential energy diagram is shown below. Use the diagram to find the maximum velocity of the particle between points A and E. At point F the particle has a speed of 14.3 m/s.

Between points A and E, at what location is the particle moving the fastest?

What is the maximum velocity achieved by the particle between points A and E? (Assume the points on the graph can be determined to three significant digits. Therefore your answer should contain at most three significant digits.)

Explanation / Answer

As the object is subject to only conservative forces , its total energy must always be constant . That is , the sum of its Potential and Kinetic Energies must be constant . So, in order to find the point where the body has the maximum velocity , we must search for the point at which the body has the least Potential energy . Clearly , that point is 'B' .

Now , it is given that the velocity of the body at point 'F' is 14.3 m/s . So , the kinetic energy = (1/2)mv2 = (1/2)*7.53*(14.3)2 = 769.90485 J . Rounding this to 3 significant digits we get , 769.905 J .

At the same point 'F' , its potential energy is 400 J . So , the total energy = 1169.905 J .

At point 'B' , potential energy is 100 J and let kinetic energy be (1/2)m(vb)2 = (1/2)*7.53*vb2 . vb = velocity at 'B' .

by conservation , 100 + (1/2)*7.53*vb2 = 1169.905 .

vb = (2 * 1069.905 / 7.53 ) 1/2 = 16.857 m/s

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