A force `F = - kx^(3)` is acting on a block moving along x-axis. Here, k is a positive constant. Work done by this force is

A force F = (10 +0.25x ) is acting on a particle moving along positive X direction . Calculate the work done by this force during a displacement from x = 1 m to x = 4 m .

The force acting on a particle moving along X-axis is F=-k(x-v_0t) where k is a positive constant. An observer moving at a constant velocity v_0 along the X-axis looks the particle . What kind of motion does he find for the particle?

The force acting on a particle moving along X-axis is F=-k(x-v_0t) where k is a positive constant. An observer moving at a constant velocity v_0 along the X-axis looks the particle . What kind of motion does he find for the particle?

The force acting on a particel moving along X-axis is F=-(x-v_0t) where k is a positive constant An observer moving at a constant velocity v_0 along the X-axis looks the prticle . What kind of motion does he find for the particle?

A force F = alpha +beta x acts on a particle of mass m along the X . Axis . Here alpha and beta are constants . Find the work done by this force when particle moves form x = 0 to x = d .

A force F=-(K)/(x^(2))(xne0) acts on a particle along the X-axis . Find the work done by the force in displacing the particale from x = +a to x = + 2a. Take K as a positive constant .

A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as F(x) = -kx + ax^(3) . Here k and a are origin as F(x) = -kx + ax^(3) . Here k and a are positive constants. For x=0 , the functional form of the potential energy U(x) of particle is.

A particle is moving along the x-axis and force acting on it is given by F=F_0sinomegaxN , where omega is a constant. The work done by the force from x=0 to x=2 will be

A particle is moving along the x-axis and force acting on it is given by F=F_0sinomega_xN , where omega is a constant. The work done by the force from x=0 to x=2 will be