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 th

Question

3 months ago

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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 th

e 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.

Physics

2 answers:

Keith_Richards [3.2K] · Answer 1 · 2026-02-23T08:32:34+03:00

Keith_Richards [3.2K]3 months ago

6 0

<span>Assuming the pulley operates without friction. Denote the tension as ‘T’. See the equation provided below.</span>

Maru [3.3K] · Answer 2 · 2026-02-23T08:31:02+03:00

The formula for the acceleration of the red block once it is released from 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 opposing the movement of the red block on the table works against the direction of motion. The motion of the overall system, including the red and grey blocks, is directed downwards. Therefore, the tension in the string is lower than the weight of 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 the red block and the table is $\mu$ .

Concept:

The total force acting on the red block follows the direction of the system's motion and equates to the tension minus the friction force.

The total force acting on the red block is:

${m_r}a = T - f$

Rearranging the above expression yields:

$T = {m_r}a + f$

Substituting $\mu {m_r}g$ for $f$ in the above equation gives:

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

Here, $T$ stands for the tension in the string, ${m_r}$ represents the mass of the red block, $a$ indicates the acceleration of the system, and $g$ is the gravitational acceleration.

The total force affecting the grey block is in the direction of the system's motion and is the difference between the weight of the grey block and the tension in the string.

The total force acting on the grey block is:

${m_g}a = {m_g}g - T$ …… (2)

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

Substituting ${m_r}a + \mu {m_r}g$ for $T$ into equation (2) allows us to write:

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

Rearranging the above expression gives:

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

Simplifying the above expression results in:

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

Learn more:

1. Change in momentum

2. The motion of a object under friction

3. A ball descending under gravitational acceleration

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

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

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 th

Response: a. The mirrors and eyepiece of a large telescope are designed with spring-loaded components to quickly return to a predetermined position.