Question

A block of mass 2 kg is connected to a freely hanging block of mass 4 kg by a light and inextensible string which
passes over pulley at the edge of a table. The 2 kg mass is on the surface of the table assumed to be smooth. Calculate
the acceleration of the system and the tension in the string

Answer 1

Answer on Question #51066, Physics, Mechanics | Kinematics | Dynamics

A block of mass $2\mathrm{kg}$ is connected to a freely hanging block of mass $4\mathrm{kg}$ by a light and inextensible string which passes over pulley at the edge of a table. The $2\mathrm{kg}$ mass is on the surface of the table assumed to be smooth. Calculate the acceleration of the system and the tension in the string.

Solution

The blocks are connected by a string. The tension due to the string is the same at both ends. According with Newton's second law

for $m_{1} = 2kg$ mass:

$$m _ {1} \vec {a} = m _ {1} \vec {g} + \vec {N} + \vec {T}$$

where $g = 10m / s^2$ is the acceleration of gravity; $\vec{N}$ is the reaction force; $\vec{T}$ is strength thread tension; $a$ is the acceleration.

for $m_{2} = 4kg$ mass:

$$m _ {2} \vec {a} = m _ {2} \vec {g} + \vec {T}$$

than

$$\left\{ \begin{array}{l} m _ {1} a = T \\ m _ {2} a = m _ {2} g - T \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = g \cdot \frac {m _ {2}}{m _ {1} + m _ {2}} = 1 0 \cdot \frac {4}{2 + 4} = 6. 6 7 m / s ^ {2} \\ T = g \cdot \frac {m _ {1} m _ {2}}{m _ {1} + m} = 1 0 \cdot \frac {2 \cdot 4}{2 + 4} = 1 3. 3 3 N \end{array} \right.$$

Answer: $a = g \cdot \frac{m_2}{m_1 + m_2} = 6.67m / s^2$; $T = g \cdot \frac{m_1m_2}{m_1 + m} = 13.33N$