# Dynamics: Acceleration of a system

1. Oct 19, 2008

### lee_s

F=ma

I found the acceleration of the system to be .1559 m/s^2
F(3kg)=(3kg)(.1559 m/s^2)=.4677N

So then I figured that the force on block M is the same as the force on the 3 kg block so,

F(3kg)=F(M) => .4677=M(g) => .4677N/9.8=M which gave me .0477kg.

I know this answer doesn't make sense but I don't know how else to approach this one.

2. Oct 19, 2008

### Stovebolt

Can you show your work for determining the acceleration of the system? I got a different answer.

I haven't run through the numbers, but it seems to me that the acceleration of the whole system times the mass of the whole system will equal the total force on the whole system. This force will be equal to the acceleration due to gravity times the mass M. This would mean you need to adjust your approach a little bit.

3. Oct 19, 2008

### lee_s

Ah! sorry about that. My first approach was I just took 1.163m/2.73057s and then took that value over the same time interval, which gave me the .1559.

But then I just tried d=vi*t+1/2*a*t^2, rearranged for acceleration and then plugged in my values:

vi*t=0 then, 2d/t^2=a and got (2)(1.163m)/(2.73057s)^2=a=.3119 m/s^2. So I'll try and run through it again with that value.

Thanks!