A 0.2 kg ball is suspended by a thread of length 1 m. It is pulled aside until the thread makes an angle of 30° with the vertical. How much work is done against gravity? The ball is now released. Find its velocity at the lowest point.

Question : A 0.2 kg ball is suspended by a thread of length 1 m. It is pulled aside until the thread makes an angle of 30° with the vertical. How much work is done against gravity? The ball is now released. Find its velocity at the lowest point. Ignore air resistance and take g = 10 m/s2. 

Doubt by Mohini 

Solution : 



m = 0.2 kg
In figure 
Length of the string = OA = OB = 1 m
In Rt. ΔOCA
cos θ = B/H = OC/OA
cos 30° = OC/1
√3/2 = OC
OC = √3/2

BC = OB - OC
BC = 1 - √3/2
h = 1-√3/2

Now, 
Work done against gravity, 
W = Fd cosθ
W = mgh cos 0°
W = mgh
W = 0.2×10×(1-√3/2)×1
W = 2-(2√3/2)
W = 2-√3
W = 2-1.732
W = 2-1.732
W = 0.268 J

Now, Applying the law of conservation of Energy

KE of Ball at Point B = PE of the Ball at Point A
½ mv² = mgh
½ v² = gh
v² =2gh
v² = 2×10×[1-(√3/2)]
v² = 20[1-(1.732/2)]
v² = 20[1-0.866]
v² = 20[0.134]
v² = 2.68
v = √(2.68)
v = 1.637 m/s

Work done by the ball against the gravity = 0.268 J
Velocity of the ball at the lowest point = 1.637 m/s