Question

A particle moves in a potential region given by
U= 8x2 - 4x + 400 J. Its state of equilibrium will be
(1) x = 25 m
(2) x = 0.25 m
(3) x = 0.025 m
(4) x = 2.5 m​

Answer 1

Explanation:

U = 8x^2 - 4x + 400

U = 8x^2 - 4x + 400Differentiate with dx we will get F

U = 8x^2 - 4x + 400Differentiate with dx we will get FF = 16x - 4

U = 8x^2 - 4x + 400Differentiate with dx we will get FF = 16x - 4At equilibrium F = 0

U = 8x^2 - 4x + 400Differentiate with dx we will get FF = 16x - 4At equilibrium F = 00 = 16x - 4

U = 8x^2 - 4x + 400Differentiate with dx we will get FF = 16x - 4At equilibrium F = 00 = 16x - 4x = 1/4 = 0.25 m

U = 8x^2 - 4x + 400Differentiate with dx we will get FF = 16x - 4At equilibrium F = 00 = 16x - 4x = 1/4 = 0.25 mIt’s state of equilibrium will be at x = 0.25 m