Question

Two masses $8\mathrm{kg}$ and $12\mathrm{kg}$ are connected at the two ends of a string that goes over a frictionless pulley. Calculate the acceleration of the masses and the tension in the string. (take, $g=10m/s^2$)

• A. $8m/s^2,144N$
• B. $4m/s^2,112N$
• C. $6m/s^2,128N$
• D. $2m/s^2,96N$

Answer 1

The correct option is D

$2m/s^2,96N$

Step-1: Given data:

Tension in the string = T

Larger mass, $m_2$ = $12\mathrm{kg}$

Smaller mass, $m_1$= $8\mathrm{kg}$

g = $10\frac{m}{s^2}$

Step-2: Calculation:

Now mass $m_2$, due to its weight, moves downwards with acceleration, and mass $m_1$ moves upwards.

Now when we apply Newton's second law of motion to the system of each mass:

Mass $m_1$ = $T-m_{1}g=m_{1}a$ …..(1)

Mass $m_2$ = $m_{2}g-T=m_{2}a$ …..(2)

When we add equations (1) and (2), we will get

$\left(m_2-m_1\right)g=\left(m_1+m_2\right)a$

$a=\frac{\left(m_2-m_1\right)g}{\left(m_1+\hspace{0.17em}{m}_2\right)}$ …..(3)

$$\begin{gathered} =\frac{(12-8)}{(12+8)}\times 10 \\ =\frac{4\times 10}{20} \end{gathered}$$

$=2m/s^2$

Here we have, the acceleration of the masses is $=2m/s^2$

Now we substitute the value in equation (2), then we will get

$m_{2}g-T=m_2\frac{\left(m_2-m_1\right)g}{\left(m_1+m_2\right)}$

$T=m_2\frac{\left(m_2^2-m_1{m}_2\right)}{\left(m_1+\hspace{0.17em}{m}_2\right)}g$

$T=\frac{2m_1{m}_{2}g}{\left(m_1+\hspace{0.17em}{m}_2\right)}$

$T=2\times 12\times 8\times \frac{10}{\left(12+8\right)}$

T = $96\mathrm{N}$

The tension in the string is $96\mathrm{N}$.