A block of mass `m_1=250 g` is at rest on a plane that makes an angle of `theta=30^@` with the horizontal. The coefficient of kinetic friction...

The speed of the second block v can be found from the equation of motion involving the distance and acceleration:

`v^2 - v_i^2 = 2ad`

Assuming the second block is falling from rest, that is,

`v_i = 0`

we get 

`v =sqrt(2ad)` ,   (Eq. 1)

where d = 30.0 cm = 0.3 m, the distance of the fall.

The main part of the problem is to find the acceleration a. To do this, we need to write down the second Newton's Law for the both blocks. Please refer to the attached image to see the direction of the forces and accelerations. The forces labeled T are the forces exerted from the spring on the blocks (that is, tension), and the force labeled N is the normal force.

`m_2vec(g) + vecT_2 = m_2veca` 

If the downward direction is taken to be positive, this becomes

`m_2g - T_2 = m_2a`       (Eq. 2)

For the first block, the second Newton's Law gives

`m_1vec(g) + vecN_1 + vecF_(f r i c t i o n) +vecT_1 = m_1veca_1 `

The component of this vector equation along the incline (upward) is

`-m_1gsin(theta) - F_(f r i c t i o n) + T_1 = m_1a_1`       (Eq. 3)

The component of this equation perpendicular to the incline is

`-m_1gcos(theta) + N_1 = 0`     (Eq. 4)

The relationship between the magnitude of the kinetic friction and the magnitude of the normal force is 

`F_(f r i c t i o n) = mu_kN_1`

Finally, note that because the string is massless and the pulley is frictionless and massless, the magnitudes of the tension forces are the same. The magnitudes of the accelerations of the blocks are also the same (otherwise, the string would stretch or break.) This means that

`T_1 = T_2 = T`

and

`a_1 = a_2 = a` .

Putting it all together, we get three equations with three variables:

(Eq. 2)               `m_2g - T = m_2a`

(Eq. 3 and 5)     `-m_1gsin(theta) - mu_kN_1 + T = m_1a`  

(Eq. 4)               `-m_1gcos(theta) + N_1 = 0`

Since we are looking for a, eliminate T and N from the equations 2 and 4, respectively. Then, we get

`-m_1gsin(theta) -mu_km_1gcos(theta) + m_2g - m_2a = m_1a`

Solving for a results in

`a = g* (m_2 - m_1(sin(theta) + mu_kcos(theta)))/(m_1 + m_2)`

Plugging in the values for the coefficient of kinetic friction, masses and angle, we get

a = 1.16 m/s^2. Notice that the result is positive for the given values, as it should be! (If it was negative, it would mean that the first block is sliding down the incline and the second block is moving up, so the equations would have to be re-written accordingly.)

Now, recall the Equation 1 which expresses the final speed of the second block in terms of acceleration:

`v = sqrt(2ad) = sqrt(2*1.16*0.3) = 0.835 m/s`

The speed of the second block is 0.835 m/s.