A child is swinging a swing. Minimum and maximum heights of swing from earths surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is:

Question

A child is swinging a swing. Minimum and maximum heights of swing from earths surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is: (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s

Tardigrade Admin · Accepted Answer

Maximum kinetic energy of swing should be equal to difference in potential energies to conserve energy.
From energy conservation

\frac{1}{2} m v_{m a x}^{2} = m g (H_{2} - H_{1})

Here,

H_{1} =

maximum height of swing from earths surface

= 0.75 m

H_{2} =

maximum height of swing from earths surface

= 2 m

∴ \frac{1}{2} m v_{m a x}^{2} = m g (2 - 0.75)

or

v_{m a x} = 2 \times 10 \times 1.25

= 25 = 5 m / s

Q. A child is swinging a swing. Minimum and maximum heights of swing from earths surface are 0.75m and 2m respectively. The maximum velocity of this swing is:

5 m/s

10 m/s

15 m/s

20 m/s

Q. A child is swinging a swing. Minimum and maximum heights of swing from earths surface are $0.75 m$ and $2 m$ respectively. The maximum velocity of this swing is: