Question

A girl in a swing is $2.5m$ above the ground at the maximum height and at $1.5m$ above the ground at the lowest point. Her maximum velocity in the swing is $\left( {g = 10m{s^{ - 2}}} \right)$
A. \[5\sqrt 2 \,m{s^{ - 1}}\]
B. \[2\sqrt 5 \,m{s^{ - 1}}\]
C. \[2\sqrt 3 \,m{s^{ - 1}}\]
D. \[3\sqrt 2 \,m{s^{ - 1}}\]
E. \[4\sqrt 2 \,m{s^{ - 1}}\]

Answer 1

Hint: Find the potential energy of the swing at the highest point and the potential energy at the lowest point. Find the difference in these potential energies. This difference in energies will be gained by the swing to attain the maximum velocity. Apply conservation of energy.

Complete step by step answer:
It is given that maximum height the girl can reach above the ground is, $h = 2.5\,m$
The minimum height the girl can reach above the ground is, $h' = 1.5\,m$
Therefore, the difference in the maximum height and the minimum height of the swing is given by, H = 2.5 m – 1.5 m = 1 m
Let m be the combined mass of the girl and the swing and v be the maximum velocity attained by the swing. Then by conservation of the energy we have
$\dfrac{1}{2}m{v^2} = mgH $
$\Rightarrow v = \sqrt {2gH} $
$\Rightarrow v = \sqrt {2 \times 10 \times 1} = 2\sqrt 5 $
Hence, the maximum velocity of the swing will be \[2\sqrt 5 \] $ms^{-1}.$

So, the correct answer is “Option B”.

Note:
According to the conservation of energy , energy can neither be created nor destroyed but it changes from one form to another. At the maximum height of the swing, the energy of the swing will be purely potential. At the lowest point of the swing it will be purely kinetic. This change in energy will provide the necessary momentum to the swing. Therefore this change in potential energy of the swing will provide the maximum velocity to the swing.