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)mv max 2=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 0.75 m and 2 m respectively. The maximum velocity of this swing is

5 ms−1

10 ms−1

15 ms−1

17 ms−1

According to law of conservation of energy 21​mvmax2​=mg(h2​−h1​) ⇒ vmax​=2g(h2​−h1​)​ =2×10×(2−0.75)​ =2×10×1.25​=25​=5ms−1

5 $m s^{- 1}$

10 $m s^{- 1}$

15 $m s^{- 1}$

17 $m s^{- 1}$

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