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A 70 kg person does a bungee jump from a bridge. The natural length of the bungee cord is 20m and it has a spring constant k of 80 N/m. When the bungee cord has a total length of 25m, what is the acceleration of the jumper (to 2 significant figures)? Hint: Draw a FBD showing the magnitudes and directions of all forces on the jumper at the instant described.

Respuesta :

Answer:

Explanation:

For this exercise we can use Newton's second law and ask us for the acceleration of the jumper

    F = ma

Let's set a reference system and take the positive direction up

    fe - W = m a

    k DX - mg = m a

we clear we calculate

     a = k / m ([tex]x_{f}[/tex] -x₀) - g

     a = 80/70 (25 -20) - 9.8

     a = 11.43 -9.8

     a = 1.6 m / s²

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