using 2565 N as the X component of tension.
... I think the place you'd lose marks here is that the reasoning behind how you arrived at that figure is not all that clear ... I see it's F
ty.tan(θ) but why would you know to do that? What was the physics behind that decision? I cannot tell from what you wrote.
hmmm. didn't think you can set FTx = 1/2 m v^2 but I'll try to remember
... this suggests that you did not get the figure the same way that Suraj did.
Do you understand why what you did got the correct answer?
Since you put in a lot of work, I'll walk you through what I mean:
The FBD should be a blob (for the mass - label it "m") with an arrow pointing directly down labelled "mg", and arrow pointing up and to one side labelled "F_T" (tension) and a vertical dotted line, the angle between the tension and the dotted line is drawn in and labelled "theta".
It is best practice to do as much algebra as possible with variables and plug the numbers in at the end.
Using ##\sum F=ma## vertically and radially off the FBD...
(1) vertical: ##F_T\cos\theta - mg = 0##
(2) horizontal: $$F_T\sin\theta = \left[ ma_{centripetal} = \frac{mv^2}{r} = \frac{mv^2}{L\sin\theta}\right]$$ ... this gives two equations with two unknowns.
Solve for the tension force in (1)...
$$F_T = \frac{mg}{\cos\theta}$$
... then sub into (2)
$$mg\tan\theta = \frac{mv^2}{L\sin\theta} $$... solve for v (L is the length of the cables: notice how the mass cancels out?): $$v=\pm\sqrt{ gL\tan\theta\sin\theta }$$
L=18m, g=9.8N/kg, θ=57deg ... plug the numbers into the equation gives v= 15.1m/s ... well done.
What I'd like you to notice is how the above approach is easier to read and troubleshoot - it is also easier to award full marks to.
Always make it easy for the marker to give you more marks.
It is often useful (and worth bonus marks) to reflect on the result. i.e. is 15m/s a fast speed to travel?
Compare with typical road speeds where you live.
