# A 2.00 kg mass is on a table attached to a string over a pulley which is then attached to another mass (1.20 kg) hanging over the edge. This is much like the set up in lab. The coefficient of...

A 2.00 kg mass is on a table attached to a string over a pulley which is then attached to another mass (1.20 kg) hanging over the edge. This is much like the set up in lab. The coefficient of kinetic friction between the table and the 2.00 kg mass is 0.200.

What will be the acceleration of the blocks?

What is the tension in the string?

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Hello!

Please look at the picture attached.

There are two forces acting on the second body: the gravitational force `m_2g` downwards and the tension force `F_T` upwards.

There are four forces acting on the first body: `m_1g` downwards, reaction force `N` upwards, tension force `F_T` (the same magnitude as for the second body) to the right and the friction force `F_f` to the left.

By the Newton's Second Law

`m_2g-F_(T)=m_2a` (the second body, projection on the downward axis)

and

`F_(T)-F_(f)=m_1a` (the first body, projection on the horizontal axis to the right).

And of course `F_(f)=mu*N=mu*m_1g.`

The magnitude of the acceleration (`a`) is the same for both bodies (the string is supposed to be inextensible).

This is a system of two equations and two unknowns (`F_T` and `a`). Express `F_T` from the first equation,

`F_(T)=m_2(g-a),`

and substitute this into the second equation:

`m_2g-m_2a-mu*m_1g=m_1a,`

or

`(m_1+m_2)a = g(m_2-mu*m_1),`

or

`a= g(m_2-mu*m_1)/(m_1+m_2) approx 9.8*(1.2-0.2*2)/(2+1.2) approx 2.45 (m/s^2).`

Therefore `F_(T)=m_2(g-a)=1.2*(9.8-2.45) approx 8.82 (N).`