A block of mass _{${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{70}\mathbf{}\mathit{k}\mathit{g}$}on a frictionless plane inclined at angle $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}$is connected by a cord over a massless, frictionless pulley to a second block of mass${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{}\mathit{k}\mathit{g}$(Figure). (a) What is the magnitude of the acceleration of each block, (b) What is the direction of the acceleration of the hanging block, and (c) What is the tension in the cord?

1. $m_1=3.70kg$
   
2. $m_2=2.30kg$
   

3.$\theta =30.0°$

From Newton’s second law, the net force on the body is equal to the vector sum of all the forces acting. The force is equal to the product of the mass of the body and its acceleration.

Apply Newton’s second law to each block to get the acceleration.

Formula:

F = ma (i)

Here, F is the net force, m is the mass of the body, and a is acceleration.

The equation of motion for block 1 along the inclined plane is

$T-m_{1}g\sin \theta =m_{1}a$

Rearranging this equation for tension T, we get

$T=m_{1}g\sin \theta +m_{1}a$ (ii)

And the equation of motion for block 2 is

$T-m_{2}g=-m_{2}a$

Rearranging this equation for tension T, we get

$T=m_{2}g-m_{2}a$ (iii)

Solving equations (ii) and (iii), we get

$a=\frac{\left(m_{2}g-m_{1}g\sin \theta \right)}{m_1+m_2}$

Substitute the values in the above equation to calculate acceleration.

$$\begin{gathered} a=\frac{\left(\left(2.30kg\right)-\left(3.70kg.\left(30°\right)\right)\right)9.8m/s^2}{\left(3.70kg+2.30kg\right)} \\ =0.735m/s^2 \end{gathered}$$

Therefore, the acceleration is$0.735m/s^2$ .

As the value of acceleration is positive, the direction of acceleration is in line with equations of motion. So, the direction of acceleration of the hanging block is downward.

We can calculate the tension using either equation (iii), we get

$$\begin{gathered} T=m_{2}g-m_{2}a \\ =\left(2.30kg\times 9.8m/s^2\right)-\left(3.70kg\times 0.735m/s^2\right) \\ =20.8N \end{gathered}$$

Therefore, the tension in the cord is$20.8N$ .