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(D) According to the work-energy theorem,the work done $(W)$ on a particle is equal to the change in its kinetic energy $(K)$.Since the particle start

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(D) According to the work-energy theorem,the work done $(W)$ on a particle is equal to the change in its kinetic energy $(K)$.Since the particle starts from rest,the initial kinetic energy is $0$.Therefore,$W = K = \frac{1}{2}mv^2$.Here,$m$ is the mass of the particle,which is constant.Thus,$W \propto v^2$.This relationship represents a parabola opening upwards along the $W$-axis.Comparing this with the given options,the graph in option $D$ represents a parabolic curve where $W$ increases with the square of $v$.

$A$ particle is initially at rest on a frictionless horizontal surface. $A$ force of constant magnitude and direction is applied to it. The graph shows the work done $(W)$ on the particle versus its speed $(v)$. If no other horizontal forces act on the particle,what should the graph look like?

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