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3 years ago

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A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

0.646. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.36 s, what is his initial speed?

1 answer:

Stels [109]3 years ago

4 0

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

$\mu$ = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N$

Magnitude of frictional force is 628.022466 N

$F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2$

$v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s$

Initial speed of the player is 8.61 m/s

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A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

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this velocity of wooden box just before going up the ramp.

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and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

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