Question

Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for the mass of the gray block, and μk for the coefficient of kinetic friction between the table and the red block. express your answer in terms of mr, mg, μk, and g.

Answer 1

Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.

 

Answer 2

The expression for the acceleration of the red block after it is released from the rest is

$\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}$

Further Explanation:

The friction force acting on the red block placed on the table is acting opposite to the direction of motion of block. The motion of the system, of red block and grey block, is vertically downward direction. It implies that the tension on the string is less than the weight on the grey block.

Given:

The mass of the red block is ${m_r}$.

The mass of the grey block is ${m_g}$.

The coefficient of friction between red block and table is $\mu$.

Concept:

The net force on the red block is acting along the direction of motion of the system and is equal to the difference between the tension in the string and friction force.

The net force on the red block is:

${m_r}a = T - f$

Rearrange the above expression.

$T = {m_r}a + f$

Substitute $\mu {m_r}g$ for $f$ in above equation.

$\fbox{\begin\\T = {m_r}a + \mu {m_r}g\end{minispace}}$                                                           …… (1)

Here,  $T$ is the tension on the string, ${m_r}$ is the mass of the red block, $a$ is the acceleration of the system and $g$ is the acceleration due to gravity.

The net force on the grey block is along the direction of motion of the system and it is equal to the difference between the weight on grey block and tension on the string.

The net force on the grey block is:

${m_g}a = {m_g}g - T$                                                                        …… (2)

Here, ${m_g}$ is the mass of the grey block.

Substitute ${m_r}a + \mu {m_r}g$ for $T$ in equation (2).

${m_g}a = {m_g}g - \left( {{m_r}a + \mu {m_r}g} \right)$

Rearrange the above expression.

${m_g}a + {m_r}a = {m_g}g - \mu {m_r}g$

Simplify the above expression.

$\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}$

Learn more:

1. Change in momentum  brainly.com/question/9484203

2.  The motion of a body under friction brainly.com/question/4033012

3. A ball falling under the acceleration due to gravity brainly.com/question/10934170

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Acceleration, friction, tension, system of two blocks, pulley mass system, motion, net force, string, block placed on the table, block hanging through a pulley, block is released from rest.