Two blocks of mass m1 = 11 kg and m2 = 13 kg are connected by a string (that doesn’t stretch) that passes over a frictionless pulley as shown in...

Hello!

The central task in such problems is to draw force diagrams correctly. After that they become not so difficult.

A. If `m_2` is accelerating down, then `m_1` is accelerating up. Their accelerations `a_1` and `a_2` (vectors) have equal magnitude `a.` The same we can say about the traction forces `F_(T1)` and `F_(T2)` and their magnitude `F_T` (from the string).

Also, if `m_1` is moving, its friction is the kinetic one.

Please look at the picture. `N_1` and `N_2` are the reaction forces, `F_(f1)` is the friction force. The friction force acts to the opposite direction to the movement (down along its side for `m_1`). It is zero for `m_2` (given).

B. Newton's Second law says that

`m_1a_1=m_1g+F_T1+N_1+F_(f1)` and

`m_2a_1=m_2g+F_T2+N_2.`

All summands are vectors here.

Also we know that

`F_T=|F_(T1)|=|F_(T2)|,` `a=|a_1|=|a_2|,` `|F_(f1)|=mu_k*|N_1|.`

To solve these equations we have to project them on some axes. The most convenient are along the side and perpendicular to that side.

C. If the string is cut. The forces `F_(T1)` and `F_(T2)` vanish and Newton's Second law equations become

`m_1a_1=m_1g+N_1+F_(f1)`  and

`m_2a_2=m_2g+N_2.`

Note that `F_(f1)` is upside the hill now (`m_1` "wants" down).

For `m_2` project its equation to the axis down along its side and obtain

`m_2a_2=m_2gsin(theta),` or `a_2=g*sin(theta) approx 6.9 (m/s^2).`

For `m_1` two projections are needed. For the downward axis along its side

`m_1a_1=m_1gsin(theta)-F_(f1),` so `a_1=gsin(theta)-F_(f1)/m_1.`

`g sin(theta) approx 6.9 m/s^2.`

For the perpendicular axis `N_1=m_1gcos(theta).`

a) If `m_1` will move down, friction will be kinetic and `F_(f1)=mu_k*N_1=mu_k m_1 g cos(theta),` so

`F_(f1)/m_1=mu_k g cos(theta) approx 0.8 (m/s^2)`

and `a_1 approx 6.9-0.8=6.1 (m/s^2).`

b) if `m_1` will be stopped manually AND released, the maximum static friction will be `m_1*mu_s*g cos(theta) approx m_1g*0.56,` which is less than `m_1g*sin(theta) approx m_1g*0.71`. So `m_1` couldn't be in rest and the a) variant is the only possible.