An 83-kg baseball player slides into second base. The coefficient of friction between the player and the ground is 0.54. If the player comes to rest after 1.7 s, what was his initial velocity?
2007-11-21 15:11:10 · 3 answers · asked by Mr. Tambourine Man 3 in Science & Mathematics ➔ Physics
The force of friction f is responcible for negative acceleration a
f=umg then
a=f/m and since
V=at we have initial velocity as
V=[umg/m]t
V=[ug]t
V=[.54 x 9.81] 1.7 (he was sliding I think a tit bit too long :-))
V=9.0 m/s (32 km/h) (That was fast)
Have fun
2007-11-21 15:17:19 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Friction=(0.54)(83)
F=ma
-(0.54)(83*9.8)=83a
a=-5.292ms^-1
V=U+at
Stops therefore final velocity = zero
0=U+-5.292(1.7)
U=8.9964ms^-1
aproxx. 9.0ms^-1
2007-11-21 23:33:45 · answer #2 · answered by Murtaza 6 · 0⤊ 0⤋
wat wat wat
2007-11-21 23:18:32 · answer #3 · answered by dude 1 · 0⤊ 0⤋