Name: Forces: Conservative and Non-conservative
Uploaded: 2019-07-19
Duration: 4 min 29 s
Description: In this video, let us define the conservative and non-conservative forces and determine which forces are conservative and which are not. Learn more.

Question

In the fig shown, a block of mass $M$ is attached to the spring and another block of mass $2 M$ has been placed over it. The system is in equilibrium. The block are pushed down so that the spring compresses further by $\frac{9 M g}{K}$ . System is released.

(a) At what height above the position of release, the block of mass $2 M$ will lose contact with the other block?

(b) What is maximum height attained by $2 M$ above the point of release?

Accepted Answer

In equilibrium $K x_{0} = 3 M g$
So spring is compressed by
$\Rightarrow x_{0} = \frac{3 M g}{K}$
Spring is further compressed by $\frac{9 M g}{K} = 3 x_{0}$

The system is released from position shown in figure $(b)$ .
Compression in the spring at this position is $4 x_{0}$ . The upper block will lose contact with the lower block when the spring acquires its natural length position.
Because now there will be retardation due to gravity and spring on $M$ mass while on $2 M$ , only gravitational retardation will act.

Hence, answer of the question is $4 x_{0} = \frac{12 M g}{k}$

Let us apply the law of conservation of mechanical energy to get the speed of the block when the spring reaches its natural length.

$\frac{1}{2} k (4 x_{0})^{2} = \frac{1}{2} 3 M V_{0}^{2} + 3 M g (4 x_{0})$ , where $x_{0} = \frac{3 M g}{K}$

$72 \frac{M^{2} g^{2}}{k} = \frac{3}{2} M V_{0}^{2} + \frac{36 M^{2} g^{2}}{k}$

$V_{0} = \sqrt{\frac{24 M}{k}} . g$

Further height attained by the block of mass $2 M$ is given by $h = \frac{v_{0}^{2}}{2 g} = \frac{12 M g}{k}$

$∴$ Height attained above point of release is $H = 4 x_{0} + h = \frac{24 M g}{k}$

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