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Starting from rest, a 98-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a magnitude of 800 N, and his speed at the bottom of the pole is 3.0 m/s. How far did he slide down the pole

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onyebuchinnaji

The distance traveled by the firefighter down the fire pole is 0.55 m.

The given parameters;

  • mass of the firefighter, m = 98 kg
  • average frictional force, F = 800 N
  • speed of the firefighter at the bottom of the ramp, v = 3 m/s.

The acceleration of the firefighter down the fire pole is calculated as follows;

[tex]-F = ma\\\\a = \frac{-F}{m} \\\\a = \frac{-800}{98} \\\\a = -8.16 \ m/s^2[/tex]

The distance traveled by the firefighter down the fire pole is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\0 = 3^2 + 2(-8.16)s\\\\16.32s = 9\\\\s = \frac{9}{16.32} \\\\s = 0.55 \ m[/tex]

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