As shown in the figure, an iron block A of volume 0 . 25   m 3 is attached to a spring S of unstretched length 1 . 0   m and hanging to the ceiling of a roof. The spring gets stretched by 0 . 2   m . This block is removed and another block B of iron of volume 0 . 75   m 3 is…

Question

As shown in the figure, an iron block A of volume 0 . 25 &#160; m 3 is attached to a spring S of unstretched length 1 . 0 &#160; m and hanging to the ceiling of a roof. The spring gets stretched by 0 . 2 &#160; m . This block is removed and another block B of iron of volume 0 . 75 &#160; m 3 is now attached to the same spring and kept on a frictionless incline plane of 30 &#176; inclination. The distance of the block from the top along the incline at equilibrium is

Aicharya · Accepted Answer

The free body diagrams of the blocks are shown in the figure below.
At equilibrium, the net force on a body is zero. Therefore,
For block $A$ ,
$m_{A} g = k x \Rightarrow k = \frac{m_{A} g}{x}$
And for block $B$ ,
$m_{B} g \sin 30 ° = k x^{'} \Rightarrow x^{'} = \frac{m_{B} g}{2 m_{A} g} x$
Since the same material is used for both the blocks, their masses will be proportional to their volume.
$\Rightarrow x^{'} = \frac{V_{B}}{2 V_{A}} x = \frac{0.75}{2 \times 0.25} \times 0.2 = 0.3 m$ .
Therefore, the distance of block $B$ from the top is $d = l + x^{'} = 1.3 m$ .

As shown in the figure, an iron block A of volume 0 . 25   m 3 is attached to a spring S of unstretched…

Solution

As shown in the figure, an iron block A of volume 0 . 25 &#160; m 3 is attached to a spring S of unstretched…

Solution

As shown in the figure, an iron block A of volume 0 . 25 m 3 is attached to a spring S of unstretched…