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A particle with mass 2.5 kg is moving in one dimension subject to a potential energy function that is a 4th-order polynomial. There are equilibrium points at x=−2 m, x=1 m, and x=3 m. The potential energy is zero at the origin. When the particle is located at x=−2 m, it has a potential energy of 19.0 J and is moving with a speed of 5.0 m/s. What is the speed of the particle when it is located at x=2 m?

User Ephesus
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Final answer:

The speed of the particle when it is located at x=2 m is approximately 6.32 m/s.

Step-by-step explanation:

The student is asking for the speed of a particle when located at x = 2 m. Given that the particle has a mass of 2.5 kg, a potential energy of 19.0 J when located at x = -2 m, and a speed of 5.0 m/s at x = -2 m, we can use the principle of conservation of mechanical energy to find the speed at x = 2 m.

First, we need to find the total mechanical energy of the particle. The total mechanical energy is the sum of the kinetic energy and the potential energy.

At x = -2 m, the potential energy is 19.0 J and the kinetic energy can be calculated using the formula: Kinetic Energy = (1/2) * mass * velocity^2. Substituting the values, we get:

Kinetic Energy = (1/2) * 2.5 kg * (5.0 m/s)^2 = 31.25 J.

Therefore, the total mechanical energy at x = -2 m is 19.0 J + 31.25 J = 50.25 J.

Since the potential energy is zero at the origin, the total mechanical energy at the origin is equal to the kinetic energy at the origin:

Total Mechanical Energy at Origin = (1/2) * mass * velocity^2 = 0.

Solving for velocity, we get:

velocity^2 = (2 * Total Mechanical Energy at Origin) / mass = (2 * 0) / 2.5 kg = 0.

Therefore, the velocity at the origin is zero.

Using the principle of conservation of mechanical energy, we can equate the total mechanical energy at x = -2 m to the total mechanical energy at x = 2 m:

Total Mechanical Energy at x = -2 m = Total Mechanical Energy at x = 2 m.

Substituting the values, we get:

19.0 J + 31.25 J = (1/2) * mass * velocity^2 at x = 2 m.

Solving for velocity, we get:

velocity^2 = (2 * (19.0 J + 31.25 J)) / 2.5 kg = 100 J / 2.5 kg = 40 m^2/s^2.

Therefore, the speed of the particle at x = 2 m is the square root of 40 m^2/s^2, which is approximately 6.32 m/s.

User Mirlande
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