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 .20 and the push imparts an initial speed of 4.0m/s?
2006-10-26 21:44:47 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
f = mew N
ma = mew mg
a = mew g
a= 1.962 m/ss
vo = 4, vf = 0, a = - 1.962m/ss
x=?
use formula
vf^2=vo^2 + 2ax
2ax = vf^2-vo^2
x= (vf^2-vo^2)/2a
x= (-16)/(2)(-1.962)
x= 4.08 m
2006-10-26 21:50:47 · answer #1 · answered by sumone^^ 3 · 0⤊ 1⤋
i am sure you could do it......
the initial kinetic energy is 1/2 m*u^2
that will have to be consumed by friction tha equals to n*N*m
for a distance s that is the desired result
do from conservation of energy law => kinetic eenrgy theorem
you have that
F*s = 1/2 *m * u^2 =>
n* N*m = 0.5*m*u^2
mass is cancelle dout therefore you end up to
s= 0.5 * u^2 / n*g
for g=10 you have that s = 0.5*4*4/0.2*10 = 4m before it reaches full stop
2006-10-26 21:53:12 · answer #2 · answered by Emmanuel P 3 · 0⤊ 0⤋
s = (v^2)/(2a)
s = (4^2)/(2*0.2*9.80662)
s = 4.0789 m
2006-10-26 21:52:57 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋