A body of mass $$0.5 \mathrm{~kg}$$ travels on straight line path with velocity $$v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}$$. The net workdone by the force during its displacement from $$x=0$$ to $$x=2 \mathrm{~m}$$ is :

In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is

A particle of mass 500 gm is moving in a straight line with velocity v = b x^5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m^$$-$$3/2 s^$$-$$1).

Arrange the four graphs in descending order of total work done; where W₁, W₂, W₃ and W₄ are the work done corresponding to figure a, b, c and d respectively.