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A body of mass 0.2 kg falls from a height of 10 m to a height of 6 m above the ground. Find the loss in potential energy taking place in the body. \[\left[ {g = 10\,{\text{m/}}{{\text{s}}^2}} \right]\]

Answer
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Hint: Use the formula for potential energy and express the potential energy of the body at its initial height and at final height. Take the difference in the potential energy to determine the change in potential energy of the body. If the difference is the negative, there is a loss in the potential energy.

Formula used:
\[U = mgh\]
Here, m is the mass of the body, g is the acceleration due to gravity and h is the height of the body above the ground.

Complete step by step answer:
We have given that the mass of the body is \[m = 0.2\,{\text{kg}}\], initial height of the body is \[{h_i} = 10\,{\text{m}}\]and the final height is \[{h_f} = 6\,{\text{m}}\].

We have the formula for gravitational potential energy,
\[U = mgh\]
Here, m is the mass of the body, g is the acceleration due to gravity and h is the height of the body above the ground.

Let’s express the change in the potential energy of the body as it falls from 10m to 6m as follows.
\[\Delta U = mg{h_f} - mg{h_i}\]
\[ \Rightarrow \Delta U = mg\left( {{h_f} - {h_i}} \right)\]

Substituting \[m = 0.2\,{\text{kg}}\], \[{h_i} = 10\,{\text{m}}\], \[{h_f} = 6\,{\text{m}}\] and \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[\Delta U = \left( {0.2} \right)\left( {10} \right)\left( {6 - 10} \right)\]
\[ \therefore \Delta U = - 8\,{\text{J}}\]
The negative sign in the above equation indicates that there is a loss in the potential energy of the body.

Therefore, the loss in the potential energy of the body is 8 J.

Note: The factor \[mg\] in the potential energy formula is the weight of the body. Always subtract the initial height from the final height of the body to determine the change in potential energy. In this way, you will be able to tell whether there is loss or gain in the potential energy by looking at the sign of the change in the potential energy.