# 2D Kinematics Problem w/ Coefficient of Friction?

1. Dec 1, 2011

### charliexx09

I'm really lost on how to do this physics problem because I don't really understand all the components of the problem:

A 7.25 kg block is sliding at 5.45m/s on a frictionless lab bench which is 8.6m high. The block hits a 1m "rough patch" before sliding off of the lab bench, and then lands 1.2m from the edge of the bench. What was the coefficient of friction between the block and the rough patch?

If you could show your work or explain how to do this it would be really helpful. Thank you! Also, this is my first post on the forums!

2. Dec 1, 2011

### Darth Frodo

Ok, this is actually a really nice question! It provides a good mix between projectiles and Newton's laws.

Here's a tip reverse engineer the projectile problem to find the initial velocity of the projectile.
Then work on the part before the rough patch!

3. Dec 1, 2011

### charliexx09

How should I determine the initial velocity of the projectile if all that is given is is the distance in x & y and the acceleration? This leaves Time, Vi & Vf unknown?

4. Dec 1, 2011

### Darth Frodo

Here's a tip, Uy = 0 m/s

U = Viy

5. Dec 1, 2011

### charliexx09

Okay, here is my process so far:
I know that d= Vi*t + (0.5)(a)(t^2), so
-8.6= -4.9t^2
t= 1.325
Then, I plugged this into the equation for the horizontal
1.2=Vi(1.325) + (-4.9)(1.325)^2
So, Vi=7.398 m/s
This is where I am really stuck seeing as I was always bad at figuring out the coefficient of friction (I've only been doing Physics for a few months!). I know that Force of Fric= u*Fnormal ... is the force of friction here the same as the initial velocity? Thanks for all your help Darth Frodo :)

6. Dec 1, 2011

### Darth Frodo

Then, I plugged this into the equation for the horizontal1.2=Vi(1.325) + (-4.9)(1.325)^2

This is wrong i'm afraid, remember there is no acceleration in the x direction.

What the answer at the back of the book?