Question

Two masses of $10\mathrm{kg}$ and $20\mathrm{kg}$ respectively are connected by a mass-less spring. A force of $200\mathrm{N}$ acts on the $20\mathrm{kg}$ mass at the instant when the $10\mathrm{kg}$ mass has an acceleration of $12\mathrm{m}/{\mathrm{s}}^2$. The acceleration of the $20\mathrm{kg}$ mass is

• A. $2\mathrm{m}/{\mathrm{s}}^2$
• B. $4\mathrm{m}/{\mathrm{s}}^2$
• C. $10\mathrm{m}/{\mathrm{s}}^2$
• D. $20\mathrm{m}/{\mathrm{s}}^2$

Answer 1

The correct option is B

$4\mathrm{m}/{\mathrm{s}}^2$

Step 1: Concept used

The system will be at equilibrium at any instant if the force acting on the system is the equilibrium.

The force acting on object 1 should be equal to the force acting on object 2.

The force acting on a body in dynamic equilibrium can be calculated by multiplying its mass by its acceleration.

The equilibrium equation can be written as follows:

$$\begin{gathered} m_1{a}_1=200\mathrm{N}-m_2{a}_2 \\ 10\mathrm{kg}\times 12\mathrm{m}/{\mathrm{s}}^2=200\mathrm{N}-20\mathrm{kg}\times a_2 \end{gathered}$$

Here,

$m_1$ and $m_2$ are the masses of the objects.

$a_1$ and $a_2$ are the accelerations of the objects respectively.

Step 2: Calculate the acceleration of the object 2

Calculate the acceleration of object 2 as follows:

$$\begin{gathered} 120\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2=200\mathrm{N}\times \left|\frac{1\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right|-20\mathrm{kg}\times a_2 \\ 20\mathrm{kg}\times a_2=80\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2 \\ {a}_2=4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the acceleration of object 2 is $4\mathrm{m}/{\mathrm{s}}^2$.

Hence, option (B) is correct.