Solving Physics Problem: 24.5N Brick on 30° Incline

[DarkAnt]

ok, this is very simple... I think. Unfortunately my physics book is useless.

A brick weighing 24.5N being released from rest on a 1.00M frictionless plane, inclined at an angle of 30 degrees. The brick slides down the incline and strikes a second brick (on a flat frictionless surface) weighing 36.8N

a) calculate the speed of the brick at the bottom of the incline.
b) if the two bricks stick together, with what initial speed will they move along?
c) If the force of friction acting on the two bricks is 5.0N, how much time will ellapse before the bricks slide before coming to rest?

I think I could do the problem myself if I knew how to do part A.

Any help would be very appreciated.

 

[Jimmy]

Here's a hint:

a = g \cdot sin(\theta)

g is the acceleration of gravity

\theta is the angle at which the plane is inclined.

 

[DarkAnt]

ty ty ty ty!

using the no time equation i can find the final velocity! thanks again

 

[DarkAnt]

uh oh

In C I have to find time it will take for the bricks to stop and I can't figure it out.

I found that Vf = 3.13m/s when it slides down and when the bricks collide the velocity of the two bricks is 1.25m/s. I found that the coefficient of friction is .08

I think I should use the equation vf = vi + at but I don't have a... or do I and I am just missing it?

could a = 5m/s^s? no it isn't that.

how about using Ff = ma? that would make a = -.08 and thus make the time = 15.6 sec. that doesn't seem correct. A little help would be nice (Jimmy come back!)

 

[Jimmy]

Sorry I had to bail on you but I had to log off for awhile to take care of personal business. I also thought you had things under control. I didn't see your last posts until now. I would have thought someone else would have helped out. <shrug>

Your method for solving problem c is correct, you just need to recalculate your acceleration. F=-5N, m=6.25Kg

 

[DarkAnt]

thanks a ton jimmy

that brings up a good question, where was everyone else?!

jk, its all good. Thanks again

 

[Jimmy]

Anytime...