A) the unstretched length of each elastic rope is 24m. The rope obeys hookes law. The vertical distance between P and Q is 35m.
Show that the total elastic potential energy stored in both ropes before the loaded cage is released is about 5*10^5J

b) The designers of the ride claim that the loaded cage will reach a height of 50m above Q. Deduce whether this claim is justified.

Explanation:

a) The rope obeys Hooke's law, so:

F = k Δx

The elastic energy in the rope is:

EE = ½ k Δx²

Or, in terms of F:

EE = ½ F Δx

Use trigonometry to find the stretched length.

cos 20° = 35 / x

x =  37.25

So the displacement is:

Δx = 37.25 − 24

Δx = 13.25

The elastic energy per rope is:

EE = ½ (3.7×10⁴ N) (13.25 m)

EE = 245,000 J

There's two ropes, so the total energy is:

2EE = 490,000 J

Rounded to one significant figure, the elastic energy is 5×10⁵ J.

b) The elastic energy in the ropes is converted to gravitational energy.

EE = PE = mgh

5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h

h = 42 m

Rounded to one significant figure, the height is 40 m.  So the claim is not justified.

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.