# Forces and friction

1. Dec 8, 2009

### joeG215

1. The problem statement, all variables and given/known data

A 1 kg box begins at rest and a hand applies a constant force of 2N along a 2 meter board. The first meter is frictionless, but the second meter has friction that makes the box come to a stop. Time is not give. Coefficient of friction is not given. My question is, is it possible to find acceleration at the 2nd meter and/or the coefficient of friction?

2. Relevant equations

$F=ma F(f)=uF(n) u= Coef. of friction 3. The attempt at a solution$F= 2N-F(f) =(1kg)(a)

2N-u(9.8N)=1kg(a) I get stuck with two variables.

2. Dec 8, 2009

### hitmeoff

Ok so for the first meter you have a box, with a force of 2N being applied to it.

F = ma, so the accel on the box is F/m = 2 m/s^2

Now this box undergoes this accel from rest through 1 meter.

v12=v02+2s(x1-x0)

v12= 0 + 2(2 m / s2)(1m)
v1 = 2m/s

Now, for the box to decelerate from 2 m/s to zero through 1m, you need to have an acceleration of -2 m/s2. You can get this from just doing another kinematic equation or by reason (if it takes 2 m/s^2 to accel from rest to 2m/s through 1 meter, then it should take the same, but opposite to decel to go from 2m/s to rest through 1 meter)

Ok, so now you know that the box must undergo a decel or 2m/s^2 to come to rest after 1 m.

This means that when you sum the forces on the box $$\sum F$$x = max = 1kg* (-2m/s^2)

The net sum must therefore be -2N

The forces acting on the box are the Fhand and the ffriction so sum Fx = Fhand - ffriction = -2N

so ffriction = 4N = $$\mu$$k*N = $$\mu$$k*mg

so $$\mu$$k = about .4

3. Dec 8, 2009

### joeG215

of course.. Thanks