Coefficient of kinetic friction and acceleration

[Leid_X09]

Homework Statement

A 3.00 kg block starts from rest and moves down a ramp on an incline of 30 degrees. If it makes it 2.00 m down the incline in 1.20 s, then:

a) what is the acceleration of the block?

This answer I got using dX=volt + 1/2at^2. The answer is 1.78 m/s^2.

b) the coefficient of kinetic friction between the block and the ramp.

Here is where I run into problems. n can be found using mgcos30 (29.4cos30) which gives you n = 25.5. But I'm confused on how to determine the frictional force. I know it can't be -fk = ma, since the object is in moving so fk should be less than ma. How can I set this equation up? What variable am I not looking at? I know that uk = Fk/n, but how do I get Fk?

 

[NBAJam100]

Your first step is going to be setting up a free body diagram. After that, keep in mind that Fk= uFn, where u= u(k)

Use Ef(x)=ma(x). you already determined the x-accel in the first part.

So from what you said, you thought it was -Fk=ma(x). So -uFn=ma(x), = -u=Ma(x)/Fn, wouldn't that value be smaller than ma?

The only problem now seems to be your negative value of u... see where you went wrong?

 

[LowlyPion]

I think you need to recalculate your acceleration. That doesn't look right.

Once you have your acceleration, what should your acceleration have been if it were frictionless? You know the component of gravity down the incline. And you have an answer for the acceleration with friction, so the difference in acceleration must be due to the frictional force acting on the mass.

Once you have that force, you have the normal component of gravity to the incline accounting for that force and the ratio is called the coefficient of friction.