Question

A pendulum consists of a 3.8 kg stone swinging on a 4.4 m string of negligible mass. The stone has a speed of 7.8 m/s when it passes its lowest point.

(a) What is the speed when the string is at 62 ˚ to the vertical?

(b) What is the greatest angle with the vertical that the string will reach during the stone's motion?

(c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

Answer 1

First, you need to find how high the string is when it is 62 degrees
cosθ = adj/hyp

to find the adjacent side, 4.4*cos 62 = 2.06 m

The height of the pendulum will be 2.06 meters above the equilibrium position. Assuming energy is conserved,

1/2mv^2= mgh + 1/2mv^2

the masses will cancel and you can solve for the second velocity

1/2*7.8^2 = 9.8*2.06 + 1/2*v^2

v = 4.52 m/s

B) assuming all energy is transferred

1/2v² = gh so h = 1/2* 7.8^2 /9.8 = 3.10 m

To find how high up it goes from equilibrium 4.4-3.10 = 1.3m

cos θ = 1.3/4.4 so θ = 72.81 degrees

C) Total energy is 1/2 mv^2 = 1/2 * 2* 7.8^2 = 60.8 J

I hope help you.....!!