Question

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

Answer 1

The potential energy of a particle of mass 1kg moving along x-axis is given by U(x)=[x^2/2-x]J. If total mechanical energy of the particle is 2J

its maximum speed = √5 m/s

Total Mechanical energy = Kinetic Energy + Potential Energy

Maximum speed will be at max kinetic energy

so Potential energy should be minimum

U(x) = x²/2  - x

differentiating both side

dU(x)/dx = 2x/2 - 1

=> dU(x)/dx =x - 1

to have minimum Potential energy

dU(x)/dx = 0

=> x -1 = 0

=> x = 1

U(x) = 1²/2 - 1 = -1/2

Total Mechanical energy = Kinetic Energy + Potential Energy

=> 2 = KE + (-1/2)

=> KE = 5/2

KE = (1/2)mv²

=> (1/2)mv² = 5/2

=> mv² = 5

m = 1 kg

=> v² = 5

=> v = √5 m/s

Answer 2

Answer:

It will help u

Explanation:

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