Question

A single conservative force F(x) acts on a 1 kg particle that moves along the x axis. The potential energy U(x) is given U(x) = 20 + (x-2)^2 joules. Where x is in meters. At x = 5m, the particle has a kinetic energy of 20 joules. The greatest value of x(in meters) to which particle can move.

Answer 1

"1kg ki jgha 2 kg, u(x) = 20 + (x-2)^2 joules, x = 5m with KE is 20 J

Fx = dU/dx = d(20 + (x-2)^2)/dx = 2 (x -2)

Hence, a = F/m = 2 (x – 2)/2 = x – 2

Velocity = v dv/dx = avdv = adx

Integrating on both sides

Vdv = integrating of x -2

V^2/2 = x^2/2 – 2x +C

V^2 = x^2- 4x + C

KE at x = 5m is 20 J

20 J = 5^2 – 2*5 + C

20 = 25 – 10 + C

20 – 15 = C

C = 5

Now, KE = x^2- 4x + 5

dK./dt = 0

2x – 4 = 0

2x = 4

X = 2

KE at x= 2 m will be

d^2k/d^2t = x

at x = 2

it will be equal to 2

hence point will be 0 and 2

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