Question

A child is swinging a swing. Minimum and maximum heights of swing from earth’s surface are 0.75 m and 2m respectively. The maximum velocity of this swing is
(a) 5 m/s (b) 10m/s (c) 15 m/s (d) 20m/s

Answer 1

Correct Answer: A

Solution :

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

12mv2max=mg(H2−H1)

Here,

H1= minimum height of swing from earth's surface = 0.75 m

H2= maximum height of swing from earth's surface = 2m

∴12mv2max=mg(2−0.75) orvmax=2×10×1.25−−−−−−−−−−−√
=25−−√=5m/s

Answer 2

Given :-

Initial height of swing = h = 0.75 m

Final height of swing = H = 2 m

From Energy Conservation,

Gain in Kinetic Energy = Loss in Potential Energy

1/2 mv² = mg(H - h)

As the mass is same so it will cancel eachother.

1/2 v² = 10(2 - 0.75)

v² = 2 × 10 × 1.25

v = √20 × 1.25

v = √25

v = 5 ms-¹

Hence,

The velocity of Swing = v = 5 ms-¹