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A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?

Respuesta :

okpalawalter8

Answer:

[tex]v=6.65m/sec[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=97.6[/tex]

Coefficient of kinetic friction  [tex]\mu k=0.555[/tex]

Generally the equation for Frictional force is mathematically given by

 [tex]F=\mu mg[/tex]

 [tex]F=0.555*97.6*9.8[/tex]

 [tex]F=531.388N[/tex]

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 [tex]a_f=-\mu g[/tex]

 [tex]a_f=-0.555 *9.81[/tex]

 [tex]a_f=-54455m/sec^2[/tex]

Therefore

 [tex]v=u-at[/tex]

 [tex]v=0+5.45*1.22[/tex]

 [tex]v=6.65m/sec[/tex]

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