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A 69.0 kg circus performer slides 3.80 m down a pole to the circus floor, starting from rest. What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of 460 N

Respuesta :

Answer:

(a) 2569.56 J

(b) 821.566 J

Explanation:

(a)When the performer slides down from 3.8m high its potential energy is converted to kinetic energy.

[tex]E_k = E_p[/tex]

[tex]E_k = mgh[/tex]

where m = 69 kg is the her mass and h = 3.8 mis the vertical distance she traveled, g = 9.8m/s2 is the constant gravitational acceleration.

[tex]E_k = 69*9.8*3.8 = 2569.56 J[/tex]

(b) If friction applies an F = 460N force on her, over 3.8m this force would have done a work of:

[tex]W_f = F * h = 460 * 3.8 = 1748 J[/tex]

Some of the potential energy will be lost to this work, so the remaining kinetic energy of her is

[tex]E_k = E_p - W = 2569.56 - 1748 = 821.56 J[/tex]