Question

A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 \[m{{s}^{-1}}\]. What will be the potential energy of the mass at the end of 2 seconds? \[(g=10m{{s}^{-2}})\]

Answer 1

Hint: We need to understand and relate the connections between the initial velocity and mass of an object which is thrown vertically upwards with the potential energy of the object with the change in time to solve this problem very easily.

Complete answer:
The potential energy of an object is the energy related to the height of the object above the earth’ surface or the nearest point of rest. For a moving object, the velocity contributes to the kinetic energy and mostly the potential energy will be absent or negligible.
In our case, an object is thrown against the gravitational force. We know that since a work is done against gravity there is a possibility that the object has a potential energy as it moves upwards. i.e., initially, when the body is thrown with a velocity, it consists only of kinetic energy. But once the object is thrown, its kinetic energy converts to potential energy gradually and becomes completely potential once it reaches the maximum height. So, we can calculate the maximum energy possessed by the object as –
\[\begin{align}
  & KE=\dfrac{1}{2}m{{v}^{2}} \\
 & \Rightarrow K{{E}_{\max }}=\dfrac{1}{2}(2){{(20)}^{2}} \\
 & \therefore K{{E}_{\max }}=400J \\
\end{align}\]
From this we can find the maximum height attained by the body is given as –
\[\begin{align}
  & P{{E}_{\max }}=K{{E}_{\max }} \\
 & \Rightarrow mgh=400J \\
 & \Rightarrow (2)(10)h=400J \\
 & \therefore h=20m \\
\end{align}\]
Using the equations of motion, we can find the height attained by the object after 2 seconds of its projection along y-axis as –
\[\begin{align}
  & S=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & \Rightarrow S=ut-\dfrac{1}{2}g{{t}^{2}} \\
 & \Rightarrow S=(20)(2)-\dfrac{1}{2}(10){{(2)}^{2}} \\
 & \therefore S=20m \\
\end{align}\]
We understand that after two seconds, the object reaches the maximum attainable height. So, the potential energy will be the maximum.
\[\therefore P{{E}_{\max }}=P{{E}_{at,t=2s}}=400J\]
This is the required solution. The potential energy will be 400J after 2 seconds.

Note:
We can understand from our discussion that the half the time of flight of the object is the time taken for the object to attain its maximum height. We can find the height attained directly using the equations of motion, but this will give a clearer idea.