(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 and Power 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 and Power

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Work Energy Theorem and Conservation of Mechanical Energy

$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

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$A$ uniform chain of mass $M$ and length $L$ is released from rest from a smooth horizontal surface as shown in the figure. The velocity of the chain at the instant it completely comes off the table will be:

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State and establish the principle of conservation of mechanical energy.

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$A$ mass of $2 \ kg$, initially at a height of $1.2 \ m$ above an uncompressed spring with spring constant $2 \times 10^4 \ N/m$, is released from rest to fall on the spring. Taking the acceleration due to gravity as $10 \ m/s^2$ and neglecting air resistance, the compression of the spring in $mm$ is:

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$A$ light rod of length $L$ is pivoted at its upper end. Two masses (each of mass $m$) are attached to the rod,one at its midpoint and the other at its free end. What horizontal velocity must be imparted to the lower end so that the rod reaches the horizontal position?

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$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

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