Question

A $\text{4 kg}$ block is connected with two springs of force constants $k_1=100{\text{Nm}}^{-1}$ and $k_2=300{\text{Nm}}^{-1}$ as shown in figure. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest postion is $(g=10{\text{ms}}^{-2})$

• A. $\text{zero}$
• B. $5{\text{ms}}^{-2}$ $\text{upwards}$
• C. $10{\text{ms}}^{-2}$ $\text{downwards}$
• D. $10{\text{ms}}^{-2}$$\text{upwards}$

Answer 1

The correct option is D $10{\text{ms}}^{-2}$$\text{upwards}$

Let $x$ be the maximum displacement of block donwards. Then from conservation of mechanical energy, we have the decrease in potential energy of $2\text{kg}$ block must be equal to the increase in elastic potential energy of both the springs. So, we have
$\text{mgx}=\frac{1}{2}(k_1+k_2)x^2$
$\Rightarrow x=\frac{2mg}{k_1+k_2}=\frac{\left(2\right)\left(4\right)\left(10\right)}{100+300}=0.2m$
Acceleration of block in this position is
$a=\frac{(k_1+k_2)x-\text{mg}}{m}$[upwards]
$\Rightarrow a=\frac{\left(400\right)\left(0.2\right)-\left(4\right)\left(10\right)}{4}=\frac{80-40}{4}$
$\Rightarrow a=10{\text{ms}}^{-2}$ [upwards]