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.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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