Posted by **Phil** on Saturday, May 5, 2012 at 9:07pm.

Two block are connected by a rope that runs over a pulley. The block on the tables has mass 4kg, the hanging block has mass 2kg, and the pulley has mass 0.5kg and radius 0.25m. Assume that the table is friction-less. If the block are released from the rest, determine their speeds after the hanging block has dropped 0.75m.

Please explain the formula and steps. Thanks.

- Physics 121 -
**Elena**, Sunday, May 6, 2012 at 12:34pm
m1 =4 kg, m2 = 2 kg, m = 0.5 kg, R = 0.25 m, h= 0.75 m.

Projections of the equation according to the 2 Newton's law for two blocks on the horizontal (for the 1st block)and on the vertical (fot the 2nd block) axis:

m1•a = T1

m2•a =m2•g-T2,

The equation of the pulley motion (2nd Nerton's law, for the rotational motion)

I•ε =M.

The moment of inertia of the pulley (disk) is

I =m•R²/2 ,

M = torque = (T1-T2)•R,

ε = a/R,

I•ε =M => m•R²•a/2•R =(T1-T2) •R =>

m•a/2 = (T1-T2).

m1•a + m2•a = T1 + m2•g -T2 = m2•g + (T1-T2) = m2•g +m•a/2,

a = m2•a/[m1+m2-m(m/2)] =

= 2•9.8/(4+2+0.125)=3.336 m/s^2,

a = v^2/2•h ,

v=sqrt(2•a•h) = sqrt(2•3.336•0.75) =

= 2.2 m/s^2

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