The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 3.0 kg, M2 = 7.0 kg, M3 = 5.0 kg.

http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype09/prob81_pulley2_3m.gif

Calculate the speed of M2 at a time 1.35 s after the system is released from rest.?

First look at M1:

F=T1-m1(g)
m1a=T1-m1(g)
a=(T1-m1(g))/m1

M2:

F=T2-T1
m2a=T2-T1
a=(T2-T1)/m2

M3:

F=m3(g)-T2
m3a=m3(g)-T2
a=(m3(g)-T2)/m3

So, you have:

a=(T1-m1(g))/m
a=(T2-T1)/m
a=(m(g)-T2)/m

From this you can solve for T1, T2, and acceleration.

To find speed:

v=at^2

You've already solved for (a), now just plug in the time given for (t).

Got it?

it did not work out with me !
I got a = 7.5
and then v = 13.66
but it's not correct !

it did not work out with me !
I got a = 7.5
and then v = 13.66
but it's not correct !

Two things:

First, the value you calculated for a is incorrect. If you use kpg0001's approach the three simultaneous equations you should be solving are:


a=\frac {(T_1-m_1g)} {m_1} \\
a=\frac {(T_2-T_1)}{m_2} \\
a=\frac {(m_3g-T_2)} {m_3}


However, there's an easier way to find a . If you consider that the sum of the external forces operating on this system that are causing it to move to the right is equal to the weight of m_2 minus the weight of m_1, then:


\Sigma F = ma \\
(m_3-m_1)g = (m_1 + m_2 + m_3) a


Now solve for a.

Second, the equation for velocity is v = at, not v = at^2.

I quit

What? 0.o

it did not work out with me !
i got a = 7.5
and then v = 13.66
but it's not correct !

Three comments:

First: Have you noticed that each number in the text of the problem is followed by some letters? They are called "units of measurements" and they are obligatory. Always! No exception! Attach the unit of measurement to each number you are posting.

Second: 7.5*1.35 is not even equal to 13.66 (providing you have used consistent units of measurement).

Third: Showing your work is not a totally bad idea.

To show your work, don't hesitate to use the answer box instead of the comment box. The latter has limited space for characters.

Can you solve it with numbers please !
It's not working out with me , I know I know it's too much being fool ! But I have no background in physics , I'm hardly doing this homework !

susus - show us where you're having trouble. In my prevoius post I suggested using:


(m_3-m_1)g = (m_1 + m_2 + m_3) a


and solve for a. What did you get?

Thanks I did it !

The answer is 2.867 m/s :)

the answer is 2.867 m/s :)

Please note that this question was posted over a year ago, so I'm sure that susus has already submitted his homework. Also, your answer is incorrect.