A ball with a mass of 230 g is tied to a light string that has a length of 2.60

Question

A ball with a mass of 230 g is tied to a light string that has a length of 2.60 m. The end of the string is tied to a hook, and the ball hangs motionless below the hook. Keeping the string taut, you move the ball back and up until the string makes an angle of 25.0° with the vertical. You then release the ball from rest, and it oscillates back and forth, pendulum style. Use g = 9.80 m/s2.
(a) If we neglect air resistance, what is the highest speed the ball achieves in its subsequent motion?
(b) Resistive forces eventually bring the system to rest. Between the time you release the ball and the time the ball comes to a permanent stop, how much work do the resistive forces do?

Explanation / Answer

a)

Firstly, find the height the ball is elevated from its equilibrium position

Applying Pythagoras' Theorem

cos 25 = (2.6-y)/2.6

y = 2.6 - (2.6*cos25) = 0.244 m

Applying conservation of energy at equilibrium position (where v is maximum -> a=0),

Change in P.E = Change in K.E

mgh - 0 = 0 - 0.5*m*v2

g*y = 0.5*v2 (m cancels out)

v = (2gy) = 2.19 m.s-1

b)

This part invloves damping, which causes amplitude of oscillation to decrease with time.

Ebefore = mgh = 0.23*9.80*0.244 = 0.550 J (ball is at release point)

Eafter = 0 J (ball is at rest)

Henceforth, work done by the resistive force should be equal to E = Ebefore

(PS:Not too sure about this part, but it makes sense for me. What's your thought on it? Cheers)

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