The coefficient of friction opposing the sliding of a 5.00 kg rubber chicken across the top of a table is 0.408. How hard must you pull to make the rubber chicken slide with an acceleration of 2.00 m/s2?

Question

anonymous · Answer

I assume that coefficient of friction is for kinetic friction, and no static friction?

Anyway, use the formula$$F=ma$$m is the mass of the chicken, and a will be 2.00

F is going to be how hard you're pulling minus the friction force:$$F=Q-N \times \mu_{k}$$So you want to solve$$P-N \times \mu_{k}=m \times a$$(N is the normal force, which is equal and opposite of gravity if the object is not moving vertically on the table).

anonymous · Answer

That Q should be a P (though it honesty doesn't matter what you call it).

anonymous · Answer

I have this, (roughly equivalent to yours)... is this correct?$$F _{NET}=ma$$
$$(F _{NET}=F _{Friction})=ma$$
$$(\mu F _{NET})=MA$$

anonymous · Answer

(.408mu*9.8)=ma

anonymous · Answer

What exactly are we solving for?

anonymous · Answer

Not quite.  You have$$F_{Friction}=ma$$But I have$$F_{You}-F_{Friction}=ma$$and you are solving for $$F_{You}$$

anonymous · Answer

By (μFNET)=MA I meant (μFN)=ma

anonymous · Answer

oh,okay!!

anonymous · Answer

So could we say that Fyou= tension?

anonymous · Answer

I suppose if you had a rope or string attached to the rubber chicken, though it isn't needed.

Why do you ask?

anonymous · Answer

Oh, in the diagram it has a string attached, and I just had to label the force

anonymous · Answer

Ahh, ok.  The only thing is that this was more of a physics question, so in future you might want to ask similar questions in that forum instead.

anonymous · Answer

Noted! Thanks a million by the way!

anonymous · Answer

You're welcome!

anonymous · Answer

One more thing, would the value we're trying to find be acceleration or velocity?

anonymous · Answer

Oh wait it would be force, wouldn't it

anonymous · Answer

Yep!  :P

anonymous · Answer

:)