A child is sitting on a swing, its minimum and maximum height from the ground are $0.75\text{m and 2m}$ respectively, its maximum velocity will be
(a). $5\text{ m/s}$
(b). $\text{8 m/s}$
(c). $10\text{ m/s}$
(d). $15\text{ m/s}$
Answer
666.9k+ views
Hint: This question is based on the concept of conservation of energy. The swing at maximum height has maximum potential energy, and we can assume that the velocity of the swing is zero at this position. The velocity will have a maximum velocity at the minimum height of the swing since total energy of the swing must be conserved.
Formula Used:
We know that the total energy of a body is the sum of its potential and kinetic energy. So, when the swing is at a maximum height we will assume that the velocity of the swing at the highest point is zero so that the total energy of the swing will be equal to its potential energy.
The kinetic energy of a body of mass m and moving with a velocity of v is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
The potential energy of a body of mass ‘m’ and at a height ‘h’ is given by,
$P.E=mgh$
Complete answer:
The total energy of the body at the maximum height is given by,
$T.{{E}_{1}}=mg{{h}_{1}}+\dfrac{1}{2}m{{v}_{1}}^{2}$
${{h}_{1}}$ is the maximum height of the swing.
${{v}_{1}}$ is the velocity of the swing at the maximum height and it is assumed to be zero.
$\therefore T.{{E}_{1}}=mg{{h}_{1}}$ .. Equation (1)
At the minimum height, the swing will have both potential kinetic energy and since the height is minimum, the potential energy will be minimum at this position and the kinetic energy will be maximum at this point.
The total energy of the body at the minimum height is given by,
$T.{{E}_{2}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$
${{h}_{2}}$ is the minimum height of the swing.
${{v}_{2}}$ is the velocity of the swing at the minimum height, which is the maximum velocity.
$\therefore T.{{E}_{2}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$ …. Equation (2)
According to the conservation of energy theorem, the total energy at the maximum height should be equal to the total energy at the minimum height, so Equation (1) and Equation (2) should be equal. So, we can write,
$mg{{h}_{1}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$
$\Rightarrow {{v}_{2}}=\sqrt{2g\left( {{h}_{1}}-{{h}_{2}} \right)}$
$\Rightarrow {{v}_{2}}=\sqrt{2\times 9.8m{{s}^{-2}}\times \left( 2-0.75 \right)m}$
$\therefore {{v}_{2}}\approx 5m{{s}^{-1}}$
So, the maximum velocity attained by the swing is $5m{{s}^{-1}}$.
So, the answer to the question is option (A).
Note: The work done by a conservative force only depends on the initial and final position of the body and not on the path taken. Gravitational force and Electrostatic force are some popular examples of the conservative force. If the work done by a force depends on the path taken by the body, then we call these forces as non-conservative forces. The frictional force is an example of a non-conservative force.
Formula Used:
We know that the total energy of a body is the sum of its potential and kinetic energy. So, when the swing is at a maximum height we will assume that the velocity of the swing at the highest point is zero so that the total energy of the swing will be equal to its potential energy.
The kinetic energy of a body of mass m and moving with a velocity of v is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
The potential energy of a body of mass ‘m’ and at a height ‘h’ is given by,
$P.E=mgh$
Complete answer:
The total energy of the body at the maximum height is given by,
$T.{{E}_{1}}=mg{{h}_{1}}+\dfrac{1}{2}m{{v}_{1}}^{2}$
${{h}_{1}}$ is the maximum height of the swing.
${{v}_{1}}$ is the velocity of the swing at the maximum height and it is assumed to be zero.
$\therefore T.{{E}_{1}}=mg{{h}_{1}}$ .. Equation (1)
At the minimum height, the swing will have both potential kinetic energy and since the height is minimum, the potential energy will be minimum at this position and the kinetic energy will be maximum at this point.
The total energy of the body at the minimum height is given by,
$T.{{E}_{2}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$
${{h}_{2}}$ is the minimum height of the swing.
${{v}_{2}}$ is the velocity of the swing at the minimum height, which is the maximum velocity.
$\therefore T.{{E}_{2}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$ …. Equation (2)
According to the conservation of energy theorem, the total energy at the maximum height should be equal to the total energy at the minimum height, so Equation (1) and Equation (2) should be equal. So, we can write,
$mg{{h}_{1}}=mg{{h}_{2}}+\dfrac{1}{2}m{{v}_{2}}^{2}$
$\Rightarrow {{v}_{2}}=\sqrt{2g\left( {{h}_{1}}-{{h}_{2}} \right)}$
$\Rightarrow {{v}_{2}}=\sqrt{2\times 9.8m{{s}^{-2}}\times \left( 2-0.75 \right)m}$
$\therefore {{v}_{2}}\approx 5m{{s}^{-1}}$
So, the maximum velocity attained by the swing is $5m{{s}^{-1}}$.
So, the answer to the question is option (A).
Note: The work done by a conservative force only depends on the initial and final position of the body and not on the path taken. Gravitational force and Electrostatic force are some popular examples of the conservative force. If the work done by a force depends on the path taken by the body, then we call these forces as non-conservative forces. The frictional force is an example of a non-conservative force.
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