The potential energy of a particle of mass 1kg moving along x axis is given by U(x) = (x^2/2-x)J. if the total mechanical energy is 2J , then find its maximum speed.?

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Answer

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Pls explain it..

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Correct answer is √5 m/s

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U=(x^2/2)-x = d^2U/dx^2=x-1 so x=1now put this value in equatn of PE u get U=-1/2 and then by work energy theorem KE+PE=2 so by this it value comes out to be √5..

Answer

Potential Energy Function:
The potential energy of a particle of mass 1kg moving along the x-axis is given by U(x) = (x^2/2 - x)J.

Total Mechanical Energy:
The total mechanical energy of the particle, denoted by E, is given as 2J.

Finding Maximum Speed:
To find the maximum speed of the particle, we can use the conservation of mechanical energy. The mechanical energy of a particle is the sum of its kinetic energy and potential energy. Mathematically, it can be represented as:

E = K + U,

where E is the total mechanical energy, K is the kinetic energy, and U is the potential energy.

Applying Conservation of Mechanical Energy:
In this case, the potential energy function U(x) is given as (x^2/2 - x)J, and the total mechanical energy E is given as 2J. Therefore, we can write:

2 = K + (x^2/2 - x).

Since the particle has mass 1kg, the kinetic energy can be written as K = (1/2)mv^2, where m is the mass and v is the velocity.

Simplifying the Equation:
Substituting the values in the equation, we have:

2 = (1/2)(1)v^2 + (x^2/2 - x).

Simplifying further:

2 = (1/2)v^2 + (x^2/2 - x).

Substituting for x:
To proceed further, we need to find the value of x. At the maximum speed, the particle will be at the equilibrium position where the net force acting on it is zero. This occurs when the potential energy function U(x) is at its minimum or maximum.

To find the minimum or maximum point, we differentiate U(x) with respect to x and equate it to zero:

dU/dx = 0.

Differentiating U(x) = (x^2/2 - x)J with respect to x, we get:

dU/dx = (d/dx)(x^2/2 - x) = (2x/2 - 1) = x - 1.

Setting x - 1 = 0, we find x = 1.

Finding Maximum Speed:
Substituting x = 1 in the equation 2 = (1/2)v^2 + (x^2/2 - x), we have:

2 = (1/2)v^2 + (1^2/2 - 1).

Simplifying further:

2 = (1/2)v^2 + (1/2 - 1).

2 = (1/2)v^2 - (1/2).

Multiplying the equation by 2 to eliminate the fraction, we get:

4 = v^2 - 1.

Adding 1 to both sides of the equation:

v^2 = 5.

Taking the square root of both sides, we have:

v = √5.

Therefore, the maximum speed of the particle is √5 m/s.

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