# Friction and wood

1. Jul 30, 2013

### Zondrina

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

A block of wood which weighs 0.72 kg is being pushed across a floor. After 2s, the block has a velocity of 1.6m/s[F]. μK = 0.64.

a) Find the net force acting on the block of wood.
b) Find the force of friction acting of the block.
c) Find the force actually pushing the block of wood.

2. Relevant equations

$m = 0.72kg$
$Δt = 2s$
$μ_K = 0.64$
$\vec{v_H} = 1.6 m/s [F]$

3. The attempt at a solution

a) Okay so I want to find the net force acting on the wood. First off, the block has no movement in the vertical direction at all so we know $F_N = F_G = ma = (0.72)(9.8) = 7.06N$ and the final vertical force $F_V = 0$.

If the block is increasing velocity over time horizontally, then it is accelerating horizontally. Using this acceleration we can find the force in the horizontal direction. I believe I can use this kinematic equation :

$\vec{v_H} = \vec{v_1} + \vec{a}Δt$
$1.6 m/s [F] = 0 + (2s) \vec{a}$ [There is no initial horizontal velocity so $\vec{v_1} = 0$]
$0.8 m/s^2 [F] = \vec{a}$

Now to find the net force which happens to be in the horizontal direction only, I use :

$\vec{F_H} = m \vec{a} = (0.72kg)(0.8 m/s^2 [F]) = 0.58N [F]$

Therefore the net force acting on the block of wood is 0.58N [F].

b) I think I need to use :

$F_K = μ_KF_N = (0.64)(7.06) = 4.52N$.

c) Logically the force acting on the wood should be the frictional force together with the net force

$F_A = F_K + F_H = 0.58N + 4.52N = 5.1N$

Therefore the block is being pushed with a force of 5.1N to get it to move forward.

2. Jul 30, 2013

### 462chevelle

you don't sseem to have a question? without doing the math it looks to be ok work. im not sure about the very last equation. but like you say it seems logical to add the forces.

3. Jul 30, 2013

### barryj

4. Jul 30, 2013

### Zondrina

The applied force $F_A$ has to exceed the frictional force for the block to be accelerating forward ( This is what I thought ).

Given that the net force in the horizontal direction is 0.58 [F] and the frictional force is 4.52N , the applied force must then be the sum of the frictional force and the net force because the applied force minus the frictional force is the net force.

EDIT : Thank you chev and barry for checking.