Question

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

Answer 1

Hint : This question can be solved by using conservation of Energy theorem. Use the concept of work done in terms of kinetic energy to solve this question. Kinetic Energy is defined as the energy possessed by an object by the virtue of its motion.

Complete step by step answer
We know that every object possesses some energy whenever it is moving.
That energy is termed as Kinetic Energy.
Also since the car is moving , it is doing some work.
Work=It is the measure of energy transfer which occurs when a force is applied on an object causing it to move to some distance.
 $ W = F \times s $
Where $ W = $ work done
 $ F = $ Force applied
 $ s = displacement $
According to law of conservation of energy can neither be created nor be destroyed. It can be transferred from one form to another.
So work done by the car is changed to its kinetic energy.
According to Work Energy theorem ,
The net work done on an object is equal to the change in its Kinetic Energy.
 $ \Rightarrow W = \Delta K $
Putting the value of work from above, we have
 $ \Rightarrow \Delta K = F \times s $
We know that
 $ F = ma $
 $ \Rightarrow \Delta K = ma \times s $
We also know that
 $ {v^2} - {u^2} = 2as $
Since $ u = 0 $
We have , $ s = \dfrac{{{v^2}}}{{2a}} $
Putting the value of s in above equation, we have
 $ \Delta K = ma \times \dfrac{{{v^2}}}{{2a}} $
 $ \Rightarrow \Delta K = \dfrac{1}{2}m{v^2} $
Hence the net work done to change is kinetic energy.
 $ \Rightarrow \Delta K = $ Final Kinetic energy – Initial Kinetic Energy
Since finally , the object is coming to rest,
So Final Kinetic Energy = $ 0 $
 $ \Rightarrow \Delta K = $ $ 0 $ – Initial Kinetic Energy
 $ \Rightarrow \Delta K = - \dfrac{1}{2}m{v^2} $
Hence the work done on the object is $ - \dfrac{1}{2}m{v^2} $ .

Note
Since the net work done is coming negative in this case , so a total amount of work = $ \dfrac{1}{2}m{v^2} $ is required to be done on the object , in order to bring the object to rest . So negative work means that work needs to be done on the object.