# Friction Force question

1. Oct 2, 2009

### taegello

Can anyone tell me why

net Force of x-component = -f(k) in a moving puck? I don't get how the net Force of x-component be just the kinetic friction force. What about the force that the puck is using to be actually moving? shouldn't that force be included in the net Force of x-component?

When an object is still on an incline,
net Force of x-component = (mg)(sin theta) - f(s) = 0

on when an object is moving, shouldn't it be something minus f(k)?

Any help would be greatly appreciated

2. Oct 2, 2009

### diazona

The puck does not "use" any force just to be moving. Once it starts moving it will continue moving on its own (as long as any friction acting on it is compensated for by another force, of course).

3. Oct 2, 2009

### taegello

but if net Fx = - f(k) the puck shouldn't be moving or at least moving with constant accelleration? isn't this saying that ma= -f(k)? if the opposing forces are the same, how can the object be moving? any force of ma will be cancelled out by -f(k)

I am so lost....

Last edited: Oct 2, 2009
4. Oct 2, 2009

### Staff: Mentor

If the net force Fx = -f(k), then the puck is accelerating.
(1) There's no opposing force. -f(k) is the only (or the net) force acting.
(2) "ma" is not a force, it's mass*acceleration. Newton's 2nd law tells us that the net force on an object will equal ma: ∑F = ma.
(3) If another force acted to cancel out the -f(k) force, then the net force would be zero, not -f(k).