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A 67.3-kg climber is scaling the vertical wall of a mountain. His safety rope is made of a material that, when stretched, behaves like a spring with a spring constant of 1.23 x 103 N/m. He accidentally slips and falls freely for 0.921 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

Sagot :

okpalawalter8

Answer:

[tex]d=0.59m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=67.3 kg[/tex]

Spring constant [tex]\mu=1.23 * 10^3 N/m[/tex]

Fall Height [tex]h=0.921m[/tex]

Generally the Energy theorem equation for momentum is mathematically given by

Change in KE=Work done by gravity + work done by spring

[tex]0=mg*(h + d) - \frac{\mu d^2}{2}[/tex]

[tex]0=(67.3 * 9.81 (0.921 + d)) -\frac{(1.23 * 10^3 * d^2}{ 2}[/tex]

[tex]0=608.1-660.213d-615d^2[/tex]

Solving Quadratic equation

[tex]d=0.59m[/tex]

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