Applications of Newton's laws (friction related)

[CAT 2]

Homework Statement

A child pushes a block of wood with a mass of 0.72 kg across a smooth table. The block starts from a position of rest and after 2 seconds its has a velocity of 1.6 m/s [forward]
The coefficient of friction is 0.64.
a) Find the net force acting on the block of wood.
b) Find the force of friction acting on the block of wood.
c) Find the force with which the child actually pushes on the block of wood.

Homework Equations

F = ma
F friction = u Fn

The Attempt at a Solution

a)
(I already calculated acceleration, it is 0.8m/s^2)

Using:
F = ma
F = (0.72kg) (0.8m/s^2)
F = 0.576N

This the net horizontal force exerted on the block of wood. b)
Then I figured that the frictional force is: (I figured that Fn = 7.06N)

F friction = u Fn
F friction = (0.64) (7.06N)
F friction = 4.52Nc) And so the applied force must be 4.52N + 0.58N = 5.1N

Correct? Thanks in advance!

 

[haruspex]

Correct?

Yes, assuming the child pushes horizontally!

 

[CAT 2]

Thanks, I have a question about it though. Why is it that Applied force = 4.52N + 0.58N? Shouldn't the friction be minused from the net horizontal force? Don't they oppose each other as the child pushes the block? Could you draw me a free-body diagram that shows which way the forces go? Or did I just miss something in the equation that would've explained this? Please answer this as I need to understand why. Thanks.

 

[haruspex]

Shouldn't the friction be minused from the net horizontal force?

No. Net force means the sum of the applied forces, using their appropriate signs.
ma = net force = push - friction

 

[CAT 2]

OK, but, aren't you saying here that friction counts against the push to equal the net force, because you are minusing friction from the push? (net force = push - friction: this is what you are saying, right?) In the problem it is added.

 

[haruspex]

OK, but, aren't you saying here that friction counts against the push to equal the net force, because you are minusing friction from the push? (net force = push - friction: this is what you are saying, right?) In the problem it is added.

No, the equation you used in post #1 is push = friction + net force, which is the same thing.

 

[CAT 2]

Alright, I think I got the equation, it's just the same equation switched up. But why is net force and friction in the same direction, are they both working against the push then? I was taught there were 4 forces working in 4 different directions, how are there two working in the same direction here?

 

[haruspex]

But why is net force and friction in the same direction, are they both working against the push then?

No. You need to distinguish between applied forces and resultant (net) force. Sum of applied = net.
Suppose one of the applied forces is F and acts in the same direction as the net force. We then have F+ sum of other applied forces = net.
If we move F across to the other side of the equation we get net - F = sum of other applied. This does not mean F is suddenly acting oppositely to the net force.

In short, mixing applied forces and net force on the same side of the initial equation can confuse. Always best to start with the form "ma = net = sum of applied" and work from there.

 

[CAT 2]

Ok, I think I got this now, thanks for all your help haruspex!