(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

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$A$ ball at rest is dropped from a height of $12 \,m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is (in $\,m$)

DifficultAP EAMCET 2013

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An object of mass $8 \ kg$ moves under the influence of a force. The position of the object varies with time as $x = \frac{1}{2} t^2$,where $x$ is in meters and $t$ is in seconds. Find the work done by the force in the first $2 \ s$ in $J$.

Medium

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$A$ car driver is trying to jump across a path as shown in the figure by driving horizontally off a cliff '$X$' at a speed of $10 \ m \ s^{-1}$. When he touches peak '$Z$' (ignore air resistance), what would be his speed (in $m \ s^{-1}$)? (use $g = 10 \ m \ s^{-2}$)

DifficultTS EAMCET 2021

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The energy required to accelerate a car from $10 \ m/s$ to $20 \ m/s$ is how many times the energy required to accelerate the same car from rest to $10 \ m/s$?

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