Respuesta :
Solution :
Let suppose [tex]$m_2g > m_1g \sin \theta$[/tex]
Then,
[tex]$a= \frac{\sum F}{\text{total mass}} $[/tex]
[tex]$=\frac{m_2g-m_1g \sin \theta - f_k}{m_1+m_2}$[/tex]
Here the frictional force, [tex]$f_k$[/tex] will be towards left on [tex]$m_1$[/tex] as the block will move towards right.
We know, [tex]$f_k = \mu_k N$[/tex] [tex]$= \mu_k m_1g \cos \theta$[/tex].
Therefore,
[tex]$a=\frac{m_2g-m_1g \sin \theta - \mu_k m_1g \cos \theta}{m_1+m_2}$[/tex]
Similarly, if [tex]$ m_1g \sin \theta > m_2g $[/tex]
Then ,
[tex]$a=\frac{m_1g \sin \theta - m_2g - \mu_k m_1g \cos \theta}{m_1+m_2}$[/tex]
And if [tex]$m_2g = m_1g \sin \theta$[/tex], then a = 0.
The acceleration of the block will be [tex]\rm a= \frac{m_1gsin\theta-m_2g- \mu_km_1gcos\theta}{m_1+m_}[/tex]. Acceleration is the rate of change of the velocity of the body.
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction
The acceleration is the ratio of force to the mass
[tex]\rm a= \frac{\sum F}{total\ mass} \\\\ \rm a= \frac{m_2g- m_1gsin\theta-\mu_k}{m_1+m_}[/tex]
The friction force is given by;
[tex]\rm f_k= \mu_k N = \mu_Km_1gcos\theta[/tex]
Hence the value of acceleration will be [tex]\rm a= \frac{m_1gsin\theta-m_2g- \mu_km_1gcos\theta}{m_1+m_}[/tex]
To learn more about the friction force refer to the link;
https://brainly.com/question/1714663
