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An 86.0-kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 × 10³ N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Respuesta :

Answer:

x = 1.95 m

Explanation:

As we know that total mechanical energy is conserved here

So gravitational potential energy of the climber will convert into spring potential energy

So we will have

[tex]mg(L + x) = \frac{1}{2}kx^2[/tex]

here we know that

m = 86 kg

L = 0.750 m

[tex]k = 1.20 \times 10^3 N/m[/tex]

now we will have

[tex]86\times 9.8(0.750 + x) = \frac{1}{2}(1.20 \times 10^3) x^2[/tex]

[tex]0.750 + x = 0.71 x^2[/tex]

[tex]x = 1.95 m[/tex]

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