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A 69-kg base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
J
(b) How far does he slide?
m

Respuesta :

The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.

we know:-

Mass = 69 kg

Speed = 3.2 m/s

Coefficient of friction ( ratio of friction force and normal force ) = 0.70

Acceleration due to gravity, g = 9.8 m/s^2

(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:

[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]

      [tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]

      [tex]= 353.28[/tex] Joules

Mechanical energy = 353.28 Joules

(b) To determine how far (distance) the runner slide:

acceleration = ug

                     [tex]= 3.2[/tex] × [tex]9.8[/tex]

                     [tex]= 31.36[/tex] [tex]m/s^{2}[/tex]

distance ,

[tex]V^{2} = U^{2} + 2aS[/tex]

[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]

[tex]10.24 = 62.72 S[/tex]

[tex]S = {\frac{10.24}{62.72} }[/tex]

Distance, S = 0.16 m  

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