A girl in a swing is 2.5 m above ground at the maximum height and at 1.5 m above the ground at the lowest point. Her maximum velocity in the swing is (g = 10 ms-2)

Question

A girl in a swing is 2.5 m above ground at the maximum height and at 1.5 m above the ground at the lowest point. Her maximum velocity in the swing is (g = 10 ms-2) (A) 5√2 ms-1 (B) 2√5 ms-1 (C) 2√3 ms-1 (D) 3√2 ms-1

Tardigrade Admin · Accepted Answer

In this case, Change in Potential energy = Kinetic energy (1/2) m v2 =m g(h2-h1) v2 =2 g(2.5-1.5) v2 =2 × 10 × 1 v =√5 × 2 × 2 v =2 √5 m / s

Q. A girl in a swing is $2.5 m$ above ground at the maximum height and at $1.5 m$ above the ground at the lowest point. Her maximum velocity in the swing is ( $g$ = $10 m s^{- 2}$ )

$52 m s^{- 1}$

$25 m s^{- 1}$

$23 m s^{- 1}$

$32 m s^{- 1}$

$42 m s^{- 1}$

In this case,
Change in Potential energy = Kinetic energy
$\frac{1}{2} m v^{2} = m g (h_{2} - h_{1})$
$v^{2} = 2 g (2.5 - 1.5)$
$v^{2} = 2 \times 10 \times 1$
$v = 5 \times 2 \times 2$
$v = 25 m / s$

Q. A girl in a swing is 2.5m above ground at the maximum height and at 1.5m above the ground at the lowest point. Her maximum velocity in the swing is (g = 10ms−2)

52​ms−1

25​ms−1

23​ms−1

32​ms−1

42​ms−1

In this case, Change in Potential energy = Kinetic energy 21​mv2=mg(h2​−h1​) v2=2g(2.5−1.5) v2=2×10×1 v=5×2×2​ v=25​m/s