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## Homework Statement

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]. μ

_{K}= 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.

## Homework Equations

##m = 0.72kg##

##Δt = 2s##

##μ_K = 0.64##

##\vec{v_H} = 1.6 m/s [F]##

## 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.