Question

Figure shown smooth curved track termination in a smooth horizontal part. A spring of spring constant $400N/m$ is attached at one end to a wedge fixed rigidly with the horizontal part. A $40g$ mass is released from rest at a height of $4.9m$ on the curved track. Find the maximum xompression of the spring.

• A. $5$
• B. $9.8$
• C. $19.6$
• D. $25$

Answer 1

The correct option is B $9.8$

The block will reach up maximum velocity (say v) when it reaches the floor and it can be found by using law of mechanical energy conservation. There will be no energy loss since the surface is smooth.

When block will collide with spring it's kinetic energy will start converting into spring potential energy and as the compression reaches maximum value (say x) it's velocity will be 0.

So we have two equations

Mgh=$\frac{M{v}^2}{2}$

$\frac{k{x}^2}{2}$=$\frac{M{v}^2}{2}$.

From these two equations we get x =9.8cm