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

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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.Since the particle is initially at rest,its initial kinetic energy is $0$.Therefore,$W = K_f - K_i = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$.Here,$m$ is the mass of the particle,which is constant.Thus,$W \propto v^2$.This relationship represents a parabola opening along the $W$-axis.Therefore,the graph between $W$ and $v$ will be parabolic in nature,as shown in option $(D)$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

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$A$ particle,initially at rest on a frictionless horizontal surface,is acted upon by a horizontal force which is constant in size and direction. $A$ graph is plotted between the work done $(W)$ on the particle and the speed of the particle $(v)$. If there are no other horizontal forces acting on the particle,the graph would look like:

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$, where $x$ is in meters and $t$ is in seconds. Find the work done in the first $6$ seconds. (in $\text{ J}$)

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$A$ particle is released from a height $H$. At a certain height $h$,its kinetic energy is two times its potential energy. The height $h$ and the speed $v$ of the particle at that instant are:

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From a height of $h$ above the ground, a ball is projected up at an angle $30^{\circ}$ with the horizontal. If the ball strikes the ground with a speed of $1.25$ times its initial speed of $40 \,ms^{-1}$, the value of $h$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,m$)

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Under the action of a force, a $2 \,kg$ body moves such that its position $x$ as a function of time $t$ is given by $x = \alpha t^2 / 2$, where $x$ is in meters, $t$ is in seconds, and $\alpha = 1 \,m/s^2$. The work done by the force in the first two seconds is: (in $\,J$)

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The position $x$ of a particle moving in one dimension under the influence of a constant force is given by $t = \sqrt{x} + 3$, where $x$ is in meters and $t$ is in seconds. Find the work done in the first $6$ seconds. (in $\text{ J}$)

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