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 ms-1 (B) 10 ms-1 (C) 15 ms-1 (D) 17 ms-1

Tardigrade Admin · Accepted Answer

According to law of conservation of energy (1/2)mv2 max =mg(h2-h1) ⇒ v max=√2g(h2-h1) =√2 × 10 × (2-0.75) =√2 × 10 × 1.25 =√25=5 ms-1

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

$5 m s^{- 1}$

$10 m s^{- 1}$

$15 m s^{- 1}$

$17 m s^{- 1}$

According to law of conservation of energy
$\frac{1}{2} m v_{m a x}^{2} = m g (h_{2} - h_{1})$
$\Rightarrow v_{m a x} = 2 g (h_{2} - h_{1})$
$= 2 \times 10 \times (2 - 0.75)$
$= 2 \times 10 \times 1.25$
$= 25 = 5 m s^{- 1}$

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

5ms−1

10ms−1

15ms−1

17ms−1