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...

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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. **

First lets set up our coordinate system. Let the x-axis follow the string and have it be +x up the ramp for block 1 and +x for block 2 be downward. We will then have the y-axes for block 1 be perpendicular from the surface of the ramp. We will not need to worry about a y-axis for block 2 since there are no other forces except those on the x-axis. Assume the string is massless and does not stretch.

We will apply Newton's second law to each block, followed by the elimination of the tension `T` and the use of the definition of the friction force `f_k` will allow us to determine the common acceleration of the blocks.

Using a constant-acceleration equation, relate the speed of the system to its acceleration and displacement

`v_x^2=v_(0x)^2+2a_x Delta x`

The initial velocity in the x direction, `v_(0x)=0`

`v_x^2=2a_x Delta x`

`eq. (1) :->` `v_x=sqrt(2a_x Delta x)`

Now apply Newton's second law, `sum F_(n e t)=ma` to block 1.

`eq. (2) :->` `sum F_x=T-f_k-F_(g,1) sin(30^@)=m_1a_x`

`eq. (3) :->` `sum F_y=F_(n,1)-F_(g,1)cos(30^@)=0`

Notice that `F_(g,1)=m_1 g` , therefore `eq. (2)` and `eq. (3)` can be written as:

`eq. (4) :->` `T-f_k-m_1g sin(30^@)-m_1g sin(30^@)=m_1a_x`

and `eq. (5) :->` `F_(n,1)=m_1g cos(30^@)`

Using `f_k=mu_kF_(n,1)` , substitute `eq. (5)` in `eq. (4)` to obtain:

`eq. (6) :->` `T-mu_km_1 cos(30^@)-m_1g sin(30^@)=m_1a_x`

Now apply `sum F_x=ma_x` to block 2.

`F_(g,2)-T=m_2a_x`

Where, `F_(g,2)=m_2a_x` , therefore:

`eq. (7) :-> m_2g-T=m_2a_x`

Add `eq. (6)` and `eq. (7)` to eliminate `T` and then solve for `a_x` to obtain:

`a_x=((m_2-mu_km_1cos(30^@)-m_1sin(30^@))g)/(m_1+m_2)`

Substituting for `a_x` in `eq. (1)` yields

`v_x=sqrt([2((m_2-mu_km_1cos(30^@)-m_1 sin(30^@))g)Delta x]/(m_1+m_2))`

Now plug in the values and simplify the results

`v_x=sqrt([2(0.200 kg-(0.250 kg)((0.100)cos(30^@)+sin(30^@)))(9.81 m/s^2)(0.300 m)]/(0.250 kg+0.200 kg))`

`v_x=84 (cm)/s`

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