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An object of mass, $m$ is moving with a constant velocity, $v.$ How much work should be done to bring the object to rest?

Last updated date: 13th Jun 2024
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Hint : To find the amount of work done to bring the object to rest we have to find the energy of the moving object. Only then we can neutralize both and bring the object to rest.

Energy of an object or substance is its ability to do work. Energy can be found in different forms. Energy has many different forms that can be grouped into two major categories namely kinetic energy and potential energy.
Potential energy is the energy stored or conserved in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance.
The kinetic energy is the energy of an object or substance that it possesses due to its motion. Having kinetic energy during its motion, the body maintains that same kinetic energy unless its speed changes. The same amount of work is done by the object when decelerating from its current speed to a state of rest.
Kinetic energy is a scalar quantity. It does not depend on direction. When we double the mass, the Kinetic energy is doubled and when we double the velocity, the kinetic energy increases by a factor of four.
Given,
An object of mass, $m$ is moving with a constant velocity, $v$
If an object is moving it possess kinetic energy
Kinetic energy $= \raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$
$m$ is the mass of the object.
$v$ is the velocity of the object.
The work done on the object to bring the object to rest is the change in kinetic energy
Change in kinetic energy → Final kinetic energy - Initial Kinetic energy
Here we take the final kinetic energy as zero because the object is brought to rest.
The initial kinetic energy of the moving object is ½ mv2
Change in kinetic energy $= {\text{ 0 - }}\raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$
Change in kinetic energy $= {\text{ }} - {\text{ }}\raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$
$- \raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$ work has to be done to bring the object to rest.

Note
The negative sign indicates we should do negative work to neutralize the kinetic energy the object possess i.e. the object possess $\raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$ energy so we should do $- \raise.5ex\hbox{$ \scriptstyle 1 $}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$ \scriptstyle 2 $} {\text{ }}m{v^2}$ to bring the object to rest.