2) A particle of mass 2/3 Kg is subjected to a potential energy functi...
Maxima and Minima Positions:
- To find the maxima and minima positions, we need to calculate the first derivative of the potential energy function V(x).
- The first derivative of V(x) is given by V'(x) = dV(x)/dx = 6x - 6x^2.
- Setting V'(x) equal to zero and solving for x, we get x = 0 and x = 1 as critical points.
- To determine if these critical points are maxima or minima, we can use the second derivative test.
- The second derivative of V(x) is given by V''(x) = d^2V(x)/dx^2 = 6 - 12x.
- Evaluating V''(0) and V''(1), we find that V''(0) = 6 > 0 and V''(1) = -6 < />
- Therefore, x = 0 corresponds to a local minimum and x = 1 corresponds to a local maximum.
Maximum Value of Potential Energy:
- To find the maximum value of the potential energy, we substitute x = 1 into the potential energy function.
- V(1) = (3*1^2 - 2*1^3) = 3 - 2 = 1 J.
- Therefore, the maximum value of the potential energy is 1 Joule.
Velocity at x = 3/2:
- To determine the velocity of the particle at x = 3/2, we need to use the conservation of mechanical energy.
- The initial mechanical energy of the particle at x = 4/3 is given by Ei = V(4/3) + Ki, where Ki is the initial kinetic energy (which we assume to be zero).
- The final mechanical energy of the particle at x = 3/2 is given by Ef = V(3/2) + Kf, where Kf is the final kinetic energy.
- Since mechanical energy is conserved, Ei = Ef. Therefore, V(4/3) = V(3/2).
- By substituting x = 4/3 and x = 3/2 into the potential energy function, we can solve for the final kinetic energy Kf.
- Once Kf is determined, we can calculate the final velocity using the equation Kf = (1/2)mv^2, where m is the mass of the particle.
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