(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

$A$ balloon filled with helium rises against gravity,increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that the density of air is constant.

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$A$ car comes to rest after traveling a distance $s$ under a constant retarding force $F$. If the mass of the car is increased by $50\%$,at what distance will the car come to rest,assuming the same retarding force $F$ and initial velocity $v$?

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$A$ spherical ball of mass $20 \, kg$ is stationary at the top of a hill of height $100 \, m$. It slides down a smooth surface to the ground,then climbs up another hill of height $30 \, m$ and finally slides down to a horizontal base at a height of $20 \, m$ above the ground. The velocity attained by the ball is ............... $m/s$.

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$A$ train engine of mass $m$ starts moving such that its velocity varies as $v = k\sqrt{s}$,where $k$ is a constant and $s$ is the distance covered. What is the total work done by all the forces acting on the engine in the first $t$ seconds?

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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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$A$ balloon filled with helium rises against gravity,increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that the density of air is constant.

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