A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope, as shown in the figure above. She starts from rest with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the woman. How high is the swinging woman above the ground level when the rope breaks? ( Hint: Use the law of conservation of energy and Newton’s Laws.)

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Respuesta :

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

[tex]T - mg cos\theta = \frac{mv^2}{R}[/tex]

[tex]T = mg cos\theta + \frac{mv^2}{R}[/tex]

now by energy conservation we have

[tex]\frac{1}{2}mv^2 = mg(R cos\theta)[/tex]

[tex]mv^2 = 2mgR cos\theta[/tex]

[tex]T = mg cos\theta + 2mgcos\theta[/tex]

[tex]T = 3mg cos\theta[/tex]

For string break down we have

[tex]T = 2mg = 3mgcos\theta[/tex]

[tex]cos\theta = \frac{2}{3}[/tex]

Now height from the ground is given as

[tex]h = R - Rcos\theta[/tex]

[tex]h = 24 - 24(\frac{2}{3})[/tex]

[tex]h = 8 cm[/tex]

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