A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.18 and the push imparts an initial speed of 4.3 m/s?
m
4.3/.18= 23. 89
2006-12-03 08:28:36 · 3 answers · asked by sleepy 1 in Science & Mathematics ➔ Physics
v^2 - v0^2 = 2as
s = (4.3^2)/(2*0.18*9.80662)
s = 5.24 m
2006-12-03 08:43:38 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
The force of friction: Ff = .18*m*g
Acceleration (actually deceleration, so call it negative):
a = -Ff/m = -.18*m*g/m
Distance:
V^2 = Vo^2 +2*a*s
where V=0, and Vo= 4.3 m/s
2006-12-03 17:22:58 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
Do you have the mass of the box? You need it, so that you can find the normal force. Then you can find the frictional force, and use that to find acceleration.
2006-12-03 16:42:06 · answer #3 · answered by Jeanne S 1 · 0⤊ 0⤋