Home
Class 12
PHYSICS
The potential energy of a 4kg particle f...

The potential energy of a `4kg` particle free to move along the x-axis varies with x according to following relationship : `U(x) = ((x^(3))/(3)-(5x^(2))/(2)+6x+3)` Joules, where `x` is in meters. If the total mechanical energy of the particle is `25.5` Joules, then the maximum speed of the particle is `x m//s`, find `x`

Promotional Banner

Similar Questions

Explore conceptually related problems

The potential energy of a 2 kg particle free to move along the x- axis is given by V(x)[(x^(4))/(4)-(x^(2))/(2)]jou l e The total mechanical energy of the particle is 0.75 J . The maximum speed of the particle ( in m//s) is :

The potential energy of a 1kg particle free to move along the x-axis is given by V(x)=(x^(4)/4-x^(2)/2)J The total mechanical energy of the particle is 2J then the maximum speed (in m//s) is

The potential energy of a 1 kg particle free to move along the x- axis is given by V(x) = ((x^(4))/(4) - x^(2)/(2)) J The total mechainical energy of the particle is 2 J . Then , the maximum speed (in m//s) is

The potential energy of a 1 kg particle free to move along the x- axis is given by V(x) = ((x^(4))/(4) - x^(2)/(2)) J The total mechainical energy of the particle is 2 J . Then , the maximum speed (in m//s) is

The potential energy of a 1 kg particle free to move along the x- axis is given by V(x) = ((x^(4))/(4) - x^(2)/(2)) J The total mechainical energy of the particle is 2 J . Then , the maximum speed (in m//s) is

The potential energy of a particle varies with x according to the reltiaon U(x) =x^(2)-4x . The point x =2 is a point of

The potential energy of a particle varies with position X according to the relation U(x) = [(X^3/3)-(3X^2/2)+2X] then