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The alarm at a fire station rings and a 84.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.47 m). Just before landing, his speed is 1.49 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

Respuesta :

Answer:

[tex]F_k = 807.82 N[/tex]

Explanation:

As we know that fireman starts from rest

so here we have

[tex]v_i = 0[/tex]

[tex]v_f = 1.49 m/s[/tex]

[tex]y = 4.47 m[/tex]

now we can use kinematics to find the acceleration

[tex]v_f^2 - v_i^2 = 2 a y[/tex]

[tex]1.49^2 - 0 = 2(a)(4.47)[/tex]

[tex]a = 0.25 m/s^2[/tex]

as we know by force equation

[tex]mg - F_k = ma[/tex]

[tex](84.5)(9.81) - F_k = 84.5(0.25)[/tex]

[tex]F_k = (84.5)(9.81 - 0.25)[/tex]

[tex]F_k = 807.82 N[/tex]

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