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 0.20 and the push imparts an initial speed of 4.0 m/s?
2006-12-26 13:33:11 · 7 answers · asked by yo yo ma 2 in Science & Mathematics ➔ Physics
F=MA
F=.2*M*g
solving for A
A=.2*g=1.96 meters/sec/sec
dV/dt=1.96
Starting at 4 meters/sec/sec
Vi=4
Finish at stopped
Vf=0
t*1.96=4
t=2.04
Since acceleration is constant
D=1/2*A*T*T=1/2*1.96*2.04*2.04
D=4.08 Meters
2006-12-26 15:29:22 · answer #1 · answered by anonimous 6 · 0⤊ 0⤋
Check it out yo.
v_f : Your final speed, which is zero, since it comes to rest due to friction.
v_i: Your initial speed, which is 4.0 m/s
a: if the deacceleration due to friction which is given by the formula a = ug, u is the coefficient of friction, and g is the gravitational acceleration of 9.89 m/s^2,
In this case your acceleration and initial velocity will has opposite signs denoting their opposition to each other. The acceleration due to friction will slow down any velocity generated by the initial push.
You can plug this all into the formula
v_f^2 = v_i^2 + 2a(x_f - x_i)
where x_f is the final location
and x_i is the initial location
and let d = (x_f - x_i) be the change in position or displacement.
Solve for d = (x_f - x_i) and that is how far it will go.
Which is about 4 meters.
Peace out.
2006-12-26 14:03:28 · answer #2 · answered by Phillip 3 · 0⤊ 0⤋
optimal Fd02c4c4cde7ae7625254d116a40f23aictid02c4c4cde7ae7625254d116a40f23anal fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a = ? x 740 = 530.8 N. Applid02c4c4cde7ae7625254d116a40f23ad fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a is 529 N. Sincd02c4c4cde7ae7625254d116a40f23a it particularly is ld02c4c4cde7ae7625254d116a40f23ass than thd02c4c4cde7ae7625254d116a40f23a optimal fd02c4c4cde7ae7625254d116a40f23aictid02c4c4cde7ae7625254d116a40f23anal fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a, thd02c4c4cde7ae7625254d116a40f23a particularly fd02c4c4cde7ae7625254d116a40f23aictid02c4c4cde7ae7625254d116a40f23anal fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a is 529N. Sincd02c4c4cde7ae7625254d116a40f23a thd02c4c4cde7ae7625254d116a40f23a applid02c4c4cde7ae7625254d116a40f23ad fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a and fd02c4c4cde7ae7625254d116a40f23aictid02c4c4cde7ae7625254d116a40f23an ad02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23a d02c4c4cde7ae7625254d116a40f23aqual and d02c4c4cde7ae7625254d116a40f23appd02c4c4cde7ae7625254d116a40f23asitd02c4c4cde7ae7625254d116a40f23a, thd02c4c4cde7ae7625254d116a40f23a nd02c4c4cde7ae7625254d116a40f23at fd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23acd02c4c4cde7ae7625254d116a40f23a appearing d02c4c4cde7ae7625254d116a40f23an thd02c4c4cde7ae7625254d116a40f23a dd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23assd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23a is d02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23a. Hd02c4c4cde7ae7625254d116a40f23ancd02c4c4cde7ae7625254d116a40f23a thd02c4c4cde7ae7625254d116a40f23a accd02c4c4cde7ae7625254d116a40f23ald02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23aatid02c4c4cde7ae7625254d116a40f23an d02c4c4cde7ae7625254d116a40f23an thd02c4c4cde7ae7625254d116a40f23a dd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23assd02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23a is [d02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23ad02c4c4cde7ae7625254d116a40f23a]0 m/s^2.
2016-12-15 08:41:16 · answer #3 · answered by pfeifer 4 · 0⤊ 0⤋
Acceleration due to friction is Cg. C: coefficient of friction.
= 0.2 x 9.8 = 1.96m/s^2.
Distance S = U^2 / 2a.
S = 4 ^2 / (2 x 1.96) = 4.08 m
2006-12-26 19:33:40 · answer #4 · answered by Pearlsawme 7 · 0⤊ 0⤋
Fnet on box = F friction
ma = (coefficient) (mg)
a = 0.2 X 9.8
a = -4.9 m/s^2
Vi = 4
Vf = 0
(Vf - Vi)/ t = 4.9
-4/4.9 = t
t = 0.816s
V avg = (Vf + Vi)/2
V avg = 2
V = d/t
d = Vt
= 2 X 0.816
= 1.63
Double check...im sure i messed up soemwhere
2006-12-26 14:11:09 · answer #5 · answered by Anonymous · 0⤊ 2⤋
a = 0.2*9.81 = 1.962 m/s²
x = V²/2a = 4²/(2*1.962) = 4.077 m
I hope the simplicity of my solution appeals!
2006-12-26 14:27:56 · answer #6 · answered by Steve 7 · 0⤊ 1⤋
this website will help you.
2006-12-26 13:43:20 · answer #7 · answered by LUNDELIUS T 2 · 0⤊ 2⤋