Acceleration in a Pulley System

[LemonBeef]

Hey, I'm doing a review packet for my AP Physics class, and I was just looking for some confirmation that I'm on the right track.

Here's the image of the problem: http://img297.imageshack.us/img297/2681/imgrx6.jpg

Ignore the scribbles. I'm using the formula a = ((mB)(g)- (µ)(mA)(g)(cos Ø)) / (mA + mB), and I'm pretty sure the answer is 2 m/s/s for both parts a and c.

Can anyone confirm/deny? Thanks.

 

[physixguru]

you are missing something...

applied force-friction=mass*accln
50-[.2*100]= 10 *a

>50-20=10*a
>30=10*a
>a=3m/s^2

this is enough for you to proceed in second case...

go ahead...

 

[Doc Al]

Ignore the scribbles. I'm using the formula a = ((mB)(g)- (µ)(mA)(g)(cos Ø)) / (mA + mB), and I'm pretty sure the answer is 2 m/s/s for both parts a and c.

Since the situations are physically different, the acceleration will be different. Don't use a canned formula* (which applies only in special cases). Instead, learn to apply first principles (force analysis and Newton's 2nd law)--then you can solve any problem tossed at you without having to memorize dozens of limited-use results.

*Does that equation even make sense for case I? There's only one mass.