Y
^ V0=4 m/sec
| N k=0.2
| /\ --->V0
| |
| |
| f<----*
-------------------|------------------>X
|
\/
mg

( I dont know why the drawing doesnt shows right at the post, it does at edit mode :/ )

N is the counter force working on the body
N=mg

1st Law of Niuton on Y axis ( no acceleration )
=============================
the sum of all forces = masa * acceleration
N-mg=0
since you havent specified the masa of the body i presume its 1
and g=9.8 round it up to 10, g=10 and since the gravity working against axis Y g=-10

N=1*-10=-10 N=-10

2nd Law of Niuton on X axis
===========================
f=ma

f=k*N=0.2N=0.2*-10=-2
-2=1*a
a=-2 // the acceleration on the x axis is -2


Now you got all you need to solve this problem

X(t)=X0+V0x*t+a*t^2/2

(a) X(t)=0+4*t-2*t^2/2

the speed equation:
V(t)=V0+ax*t
when the body will stop its speed will be 0
V(t)=0
0=4-2*t
t=2sec // after 2 sec the body will stop
now we put t=2sec in equation (a)
X(t=2sec)=4*2-2*4/2=8-4=4 meters

So the bodey will stop after 4 meters

[edited by - spree on December 5, 2003 1:48:52 AM]