Answer with Explanation:
We are given that mass of block=0.0600 kg
Initial speed of block=0.63 m/s
Distance of block from the hole when the block is revolved=0.47 m
Final speed=3.29 m/s
Distance of block from the hole when the block is revolved=
a.We have to find the tension in the cord in the original situation when the block has speed =

Because tension is equal to centripetal force
Substitute the values

b.

c.Work don=Final K.E-Initial K.E



Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)
Since you know the acceleration is 12 m/s^2, the initial velocity is 2.39*10^4 m/s and the time (you have to convert to seconds) is 5400 seconds, then you can use the equation
v = vo + at
When you plug in the values you get
v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.