The potential energy (in Joules) of a particle of mass 12.5 kg is given by the function

\[ U(x) = 1500 - 70x^2 + x^4 \]

The particle can travel only along the x-axis and is under the influence of a conservative force. The particle is released at 5 m with a speed of 6 m/s.

A. Determine the total mechanical energy of the particle.

\[ E_{\text{total}} = \text{ J} \]

B. What is the speed of the particle at $ x = 7 $ m?

\[ v = \text{ m/s} \]

C. Determine the two turning points of the motion of the particle. Enter your answer such that $ x $ represents the turning point with a larger numerical value (i.e., $ x_1 > x_2 $).

To determine the particle's total mechanical energy, we add its potential and kinetic energies. For the speed at a certain point, we use the conservation of energy, and for the turning points, all the energy will be potential at those points.

Explanation:

Based on the information of the problem, we need to calculate total mechanical energy (Potential and Kinetic) and motion turning points. The total mechanical energy of the particle will be given by the sum of its potential energy and its kinetic energy. The potential energy is calculated using the function U(x) = 1500−70x²+x⁴, and we substitute x=5m into this, giving us the potential energy at the given point.

The kinetic energy is calculated using the formula K = 0.5 * mass * velocity², substituting the given mass (12.5 kg) and velocity (6 m/s), giving us the kinetic energy at that point. For part B, we use conservation of mechanical energy law where the sum of kinetic and potential energy at x=5m must be equal to the sum at x=7m.

Now, for part C, we need to find the turning points. These are points where the velocity is zero, implying that all the energy is potential. We equate the total mechanical energy to the potential energy and solve for x. The roots of this equation will give us the two turning points.

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