Question

A boy is swinging on a swing such that his lowest and highest positions are at heights of 2m and 4.5m respectively. His velocity at the lowest position is
A) $2.5m{s^{ - 1}}$
B) $7m{s^{ - 1}}$
C) $14m{s^{ - 1}}$
D) $20m{s^{ - 1}}$

Answer 1

Hint: The total mechanical energy in a system remains constant throughout the motion. Swinging is a harmonic motion where at the highest point it has the tendency to swing in the opposite direction with the least velocity and in the lowest position it has the least potential energy with the maximum velocity.

Formula Used:
The potential energy $P$ of an object with mass $m$ at a height $h$ can be defined as
$P = mgh$
where, $g$ is the acceleration due to gravity.
The kinetic energy $K$ of an object with mass $m$ with a velocity $v$ can be defined as
$K = \dfrac{1}{2}m{v^2}$

Complete step by step answer:
The highest position of the swing is $4.5m$.
The highest position of the swing is $2m$.
We need to find the boy’s velocity at the lowest point.

Step 1:
Let the mass of the boy is $M$
The boy swings from the lowest height to the highest height which provides it the effective height $h$
$h = \left( {4.5 - 2} \right)m = 2.5m$
Calculate the potential energy ${P_{highest}}$ of the boy at the highest point.
${P_{highest}} = mgh$
At the highest height, the boy acquires no velocity.
Hence calculate the kinetic energy at that point.
${K_{highest}} = \dfrac{1}{2}m \times 0 = 0$

Step 2:
At the lowest point, the height of the boy becomes zero from the lowest point
Calculate the potential energy ${P_{lowest}}$ of the boy at the highest point.
${P_{lowest}} = mg \times 0 = 0$
At the lowest height, the boy acquires maximum velocity, say ${v_{lowest}}$.
Hence calculate the kinetic energy at that point.
${K_{lowest}} = \dfrac{1}{2}mv_{lowest}^2$

Step 3:
The total mechanical energy at any instance of a journey is always conserved.
$
{P_{highest}} + {K_{highest}} = {P_{lowest}} + {K_{lowest}} \\
\Rightarrow mg\left( {4.5 - 2} \right) + 0 = 0 + \dfrac{1}{2}mv_{lowest}^2 \\
\Rightarrow 2.5g = \dfrac{1}{2}v_{lowest}^2 \\
\Rightarrow {v_{lowest}} = \sqrt {5 \times 9.8} = \sqrt {49} = 7 \\
\therefore {v_{lowest}} = 7m{s^{ - 1}} \\
 $

If a boy is swinging on a swing such that his lowest and highest positions are at heights of $2m$ and $4.5m$ respectively, then his velocity at the lowest position is $7m{s^{ - 1}}$. So, option (B) is correct.

Note:
You can argue the same in a different way. You can see that the potential energy at the highest point is completely transferred into the kinetic energy of the system. So thus you can get the same relation and answer easily. For calculating the potential energy, the only potential difference is to be calculated. So you can take any point as the reference base which is here taken as by the lowest point where the boy starts swinging.