Recent content by nperk7288

Not sure if I did what you said correctly or not but here's what I've gotten: mgh = 1/2(Iω2) + 1/2(mv2) Then by substituting in v/r = ω I got v = 5.66 m/s After that I found the acceleration to be 5.34 m/s2 and found the Force downward to be 32 N. Since the force without the rope would...

I just took my Physics Midterm and there was one problem on the test that I had no clue on and just left blank. Here's the problem: A 6.0 kg box is attached to a rope that is wound around a pulley of radius 0.150 m and mass 10.0 kg. The box is initially at rest and is suspended 3.0 m above...

I am still lost...

ok that would be FN + Fmg how can I figure that out without knowing the mass of the box...?

But we don't know the normal force

So 47.0 N = fs right? but I don't know what N is...

How does that give me the coefficient though?

That equals fs + FP

I am still completely lost... I really don't understand this... I'm about to break down!

There is no force applied to the lower block in the first part. Only to top block is moving. The problem doesn't want to slide the top block back, the person just resets it manually and then wants to slide the lower one... -------------------- ---------> | Box 1 |...

I know the laws but that has to do with an object staying at rest or in motion unless acted on by a net force

How does Newton's frist law help to get the coefficient of Friction?

I really need help quick; I still have a big lab to do for tomorrow as well so I will have to leave to do that in a couple minutes.

For #1 I had 4 forces acting on the top box, FN pointing in the positive Y direction, FW or the Weight pointing in the negative Y direction, fs (the static friction force) pointing in the negative X direction, and FP = 47.0 N pointing in the positive X direction. On the bottom box I have 6...

I'm really struggling in Physics right now because I'm doing an 8-unit full year course in 8 weeks...It's hard to understand everything when we go as quick as we have been.